Post on 31-Jan-2020
Bending Deflection
mi@seu.edu.cn
• The Elastic Curve, Deflection & Slope （挠曲线、挠度和转角）
• Differential Equation of the Elastic Curve（挠曲线微分方程）
• Deflection & Slope by Integration（积分法求挠度和转角）
• Boundary Conditions（边界条件）
• Symmetry Conditions（对称性条件）
• Continuity Conditions（连续性条件）
• Direct Integration from Distributed Loads（直接由分布荷载积分求挠度和转角）
• Direct Integration from Transverse Loads（直接由剪力积分求挠度和转角）
• Deformations in a Transverse Cross Section（梁横截面内的变形）
• Curvature Shortening（梁由于弯曲造成的轴向位移）
Contents
2
• Deflection & Slope by Superposition（叠加法求挠度和转角）
• Superposition of Loads（荷载叠加法）
• Superposition of Rigidized Structures（刚化叠加法）
• Combined Superposition（荷载和变形组合叠加法）
• Deflection & Slope by Singular Functions（奇异函数法求挠度和转角）
• Deflection & Slope by Moment-Area Theorems（图乘法求挠度和转角）
• Stiffness Condition（刚度条件）
• Ways to Increase Flexural Rigidity（梁的刚度优化设计）
• Bending Strain Energy（弯曲应变能）
Contents
3
• The elastic curve: beam axis under bending, required to
determine beam deflection and slope.
x
w
w
Deflection curve
The Elastic Curve, Deflection and Slope
• Bending deflections (w = f(x)): vertical deflection of the neutral
surface, defined as downward positive / upward negative.
• Slope (θ = θ(x) ≈ tan(θ) = dw/dx): rotation of cross-sections,
defined as clockwise positive / counter clockwise negative
4
• Curvature of the neutral surface
zEI
xM
x
)(
)(
12 3/2
1
( ) (1 )
ww
x w
EIw M
w
x
M M
M 0
0w
w
x
M 0
0wM M
EI: flexural rigidity
Differential Equation of the Elastic Curve
5
• The negative sign is due to the particular choice of the w-axis.
)(xMwEI
CxxMwEI d)(
DCxxxxMEIw dd)(
• Integration constants C and D can be determined from boundary
conditions, symmetry conditions, and continuity conditions.
Deflection and Slope by Integration
• Conventionally assuming constant flexural rigidity (EI)
6
0; 0A Bw w
Boundary Conditions – Simple Beams
• Deflections are restrained at the hinged/rolled supports
7
0; 0A Aw
8
Boundary Conditions- Cantilever Beams
• Both the deflection and rotation are restrained at the
clamped end
Symmetry Conditions
• Both the geometry and
loads are symmetric about
the mid-section (x = L/2)
0C
9
Continuity Conditions
1 2 3
1 2 1 2
1 3 1 3
0 , , 0
;
;
x a a x L x L a
w x a w x a x a x a
w x a w x L a x a x L a
10
P
Direct Integration from Distributed Loads
• Equation for beam displacement
becomes
4 2
4 2
d w d MEI q x
dx dx
43
2
2213
161 CxCxCxC
dxxqdxdxdxxwEI
• Integrating four times yields
• For a beam subjected to distributed loads
• Constants are determined from
conditions on the shear forces and
bending moments as well as conditions
on the slopes and deflections.
2
SS 2
,dFdM d M
F x q xdx dx dx
11
Direct Integration from Transverse Loads
• Equation for beam displacement
becomes
xFdx
dM
dx
wdEI S
3
3
43
2
221
S
CxCxC
dxxFdxdxxwEI
• Integrating three times yields
• For a beam subjected to transverse loads
(without distributed loads)
• Constants are determined from
conditions on the bending moments as
well as conditions on the slopes and
deflections. 12
Deformations in a Transverse Cross Section
1 x xy y
y Ey
,y x z x
y yy y y y
1 anticlastic curvature
z y
y
13
• Deformation due to bending moment is
quantified by the curvature of the neutral surface
• Although cross sectional planes remain planar
when subjected to bending moments, in-plane
deformations are nonzero,
• For a rectangular cross-section, no change in the
vertical dimension will be observed.
• Horizontal expansion above the neutral surface
and contraction below it cause an in-plane
curvature
Curvature Shortening
2 2
2
0
11 1
21
2
L
AB AB
ds dx w dx w dx
L L w dx
• When a beam is bent, the ends of
the beam move closer together.
• It is common practice to disregard
these longitudinal displacements.
• For immovable supports, a
horizontal reaction will
develop at each end.
HL EA H EA L
• This equation gives a close estimate of the tensile stress produced by
the immovable supports of a simple beam.14
• Given: flexural rigidity (EI) of a simply supported beam under a
uniformly distributed load of density q
• Find: equations of deflections and slopes, and their maximum values
(θmax, wmax)
q
l
Sample Problem
15
• Solution:
M xql
xq
x( ) 2 2
2
2
22x
qx
qlwEI
Cxq
xql
wEI 32
64
DCxxq
xql
EIw 43
2412• Boundary conditions: 0 0, 0w x w x l
3
, 024
qlC D
16
x
q
l
w
AB
• Equations of beam deflection and slope
• The maximum deflection and slope
)2(24
323 xlxlEI
qxw
)46(24
323 xlxlEI
q＋－
x
q
l
w
AB
3
max24
A B
ql
EI
4
max
5
2 384
l qlw w x
EI
17
x
q
l
AB
w
P
xBA
l
• Given: flexural rigidity (EI) of a cantilever beam under a
concentrated load acting at its free end
• Find: equations of deflections and slopes, and their maximum values
(θmax, wmax)
Sample Problem
18
• Solution:
M x P l x( ) ( )
lPxPwEI
CxlPxP
wEI 2
2
DCxxlP
xP
EIw 23
26
• Boundary conditions: 0 0, 0 0w x w x
0C D
19
w
P
xBA
lx
)2(2
xlEI
xP
)3(6
2
xlEI
xPw
• The maximum deflection and slope
EI
PlB
2
2
max
EI
Plww B
3
3
max
• Equations of beam deflection and slope
20
w
P
xBA
lx
• Given: a simply supported beam with flexural rigidity EI is subjected
to a concentrated load P as shown
• Find: the equations of deflection and slope, and their maximum
values (wmax, θmax)
wl
2
l
2
xA B
C
P
Sample Problem
21
• Solution
2
3
( ) , 02 2
2
4
12
P lM x x x
PEIw x
PEIw x C
PEIw x Cx D
• Left boundary condition: 0 0 0w x D
• Because of symmetry, it’s sufficient to solve only portion AC.
• Symmetry condition:
22
2
0 2 16
l Plw x C
x
wl
2
l
2
xA B
C
P
• Equations of bending deflection and slope:
2 2
2 2
( 4 )16
(3 4 )48
Pl x
EIPx
w l xEI
• Maximum deflection
and slope:
2
max16
A B
Pl
EI
3
max
248
lx
Plw w
EI
23
x
wl
2
l
2
xA B
C
P
• Given: a simply supported beam with flexural rigidity EI is subjected
to a uniformly distributed load with density q, on its central portion
as shown
• Find: the equations of deflection and slope, and their maximum
values (wmax, θmax).
w
x
a a a a
A BC D E
q
Sample Problem
24
• Solution
1 1 1 1
2
2 2 2 2 2
1 1
2
2 2 2
( ) (0 )
( ) ( ) ( 2 )2
( )2
M x qax x a
qM x qax x a a x a
EIw qax
qEIw qax x a
w
Ax
a a a a
Cx1
x2
BD E
q
qaqa
• Thanks to symmetry, it is sufficient to consider only the left half
25
• Continuity conditions:
1 1 2 2 1 2 1 2 , w x a w x a C C D D
2
1 1 1
1 13
1 1 1 1 1
2
6
qaEIw x C
EIw qaxqa
EIw x C x D
• Constraint condition:
• Due to symmetry:
1 1 10 0 0w x D
3
2 2 2
112 0
6w x a C qa
2
2 2 2
2 3
2 2 2 2
3 4
2 2 2 2 2 2
( )2
( )2 6
( )6 24
qEIw qax x a
qa qEIw x x a C
qa qEIw x x a C x D
26
• Equations of deflection and slope:
axaxaaxaxEI
qw
axxxaEI
qaw
axaaaxaxEI
q
axxaEI
qa
2]44)(4[24
0)11(6
2]11)(3[6
0)311(6
22
34
2
3
22
1
3
11
2
1
2
33
2
2
22
1
2
1
2
1
• Maximum deflection and slope:
1 2
3 4
max 1 0 max 2 2
11 19
6 8A x x a
qa qaw w
EI EI ，
27
- Beam material obeys linearly elastic Hooke’s law.
• Deformation of beams subjected to combinations of loads may be
obtained as the linear combination of the deformations due to
individual loads.
Deflection and Slope by Superposition
• Superposition of Loads:
- No interactions exist among deformations induced by
individual loads.
- Procedure is facilitated by tables of solutions for common
types of loadings and supports.
28
• Using method of superposition to find the deflection at section C
and the slopes at sections A and B.
Sample Problem
qPm
AC
B
2l
2l
29
4 3 2
3 2
3 2
5
384 48 16
24 16 3
24 16 6
C
A
B
q l P l mlw
EI EI EIq l P l ml
EI EI EIq l P l ml
EI EI EI
• Superpose the deformations due to the uniformly distributed load (q),
the concentrated load (P) and the concentrated moment (m).
• Solution:
qPm
AC
B
2l
2l
q
A B
P
AC
B
m
A B
30
• Find the deflections at sections C and D.
Sample Problem
q
qD
A BC
2aaa
31
• Solution
4 45 (2 ) 50,
384 24C D
q a qaw w
EI EI
q
q
DA B
C
2aaa
q
q
DA B
C
2aaa
||
32
• Find the deflection at section C and the slope at section B.
A BC
l /2
q
l /2
Sample Problem
33
• Solution
l /2 l /2
A BC
q
A B
q/2
A B
q/2
q/2
3
4
2
2
24
52
384
B
lC x
ql
EIq
l
w wEI
3
2 2
240
B
C
q l
EIw
34
• Given θB = 0, determine the relationship between m and P.
Sample Problem
Pm
BCA
a a
35
• Solution:
2 20
2
4
B
Pa m a
EI EI
Pam
Pm
BCA
a a
P
BCA
a a
m
BCA
2a
36
• Applicable to multi-span beams
Deflection and Slope by Superposition
• Superposition of Rigidized Structures:
• The total deflection of a multi-span beam under a given loading
condition can be determined by superposing several beams
corresponding to rigidizing all but one span of the beam, under the
exactly same loading condition as the original beam.
37
• Find the deflection at section C of the simply supported
overhanging beam shown.• Solution
L a
CA
B
P
• Deflection at C due to
rigidization of portion AB
3
13
c
Paw
EI
EI
Pac
2
2
1
AB
C
P
Rigidization
EI =
Sample Problem
38
CAB
P
Rigidization
EI=
2 23
c B
PaLw a a
EI 2 2
3C B
paL
EI
1 2
23
3 3 3
c c cw w w
Pa a LPa PaLa
EI EI EI
1 22
2 3 2 3
c c c
Pa PaL Pa a L
EI EI EI
• Deflection at C due to rigidization of portion BC
• Total deflection and slope at C:
PPa
39
• Given wC = 0, determine the relationship between P and q.
Superposition of Loads & Rigidized Structures
P
DB
CA
q
a a a
40
• Solution:
4 25 (2 ) (2 )0
384 16
5
6
C
q a Pa aw
EI EI
P qa
P
DB
CA
q
a a a
DB
CA
q
a a a
P
DB
CA
a a a
P
DBCA
a a a
Pa
41
• Using the method of superposition find the deflection and slope at
section C of the beam shown.
Sample Problem
qa q
aaa
A B
C
42
3 3 3 4 4 45,
12 6 4 12 8 24C C
qa qa qa qa qa qaw
EI EI EI EI EI EI
2
2 3
4
2(2 )2
3 16 12
12
C B
C B
qaa
qa a qa
EI EI EIqa
w aEI
• Solution:
• Rigidizing BC
• Rigidizing AB
3 4
,6 8
C C
qa qaw
EI EI
• Total:
qaqa
A B
C
2 2qa
CB
q
43
• A stepped cantilever, as shown, is subjected to a concentrated load F
at its free end. Find the deflection at the free end.
F
AB
Cl/2 l/2
EI2EI
Sample Problem
44
AB C
F
l/2 l/2
EI 2EI
• Solution
FB
AwA1
θA1
• Rigidizing section BC makes AB
a cantilever subjected to a
concentrated load at its free end.
F
BAwB
C
2B l Fl/2
• Rigidizing section AB makes
the whole beam a cantilever.
1 2
1 2
33
16
A A A
lA B B
w w w
w w
Pl
EI
45
• Find the deflections at sections B and D of the beam shown below.
Sample Problem
q 2qa
aa2a
A B DC
46
• Solution
4 3 4
4
(2 ) (2 ) 14
8 3 3
7
2 3
B
BD
q a qa a qaw
EI EI EI
w qaw
EI
• Rigidizing AB
32 (2 )0
48B D
qa aw w
EI ，
• Rigidizing BC
4 4 3 414 7 2 (2 ) 8
3 3 48 3B D
qa qa qa a qaw w
EI EI EI EI ，• Total:
q2qa
aa2a
B DC
A
2qa
CDBA
q
A
q
B
2qa
D C
qa
47
• For the structure composed of a beam and a frame shown, find the
deflection at the center of the beam AB.
lA
B
l/2
E
F
lEI
EA
EIEI C
D
l/2
• The deflection at section E is
associated with the following
deformations:
• Bending of beam AB itself.
• Bending of BC
• Compression and bending of CD
• Rigidize the frame (BC+CD)F
A BE3
148
E
Flw
EI
• Solution
Sample Problem
48
• Rigidize AB + CD3 3
22 1
1 1 ( )
2 2 3 12
F
E B
l Flw w
EI EI
F/2
CBwB1
• Rigidize AB + BC
)(2
1323 BBE www
EA
FlwB
22
(Deflection at B due to the compression of CD)
3
23
( )
2
F
B C
Fll lw l l
EI EI
(Deflection at B due to the bending of CD)
F/2
D
wB2
wB3
C
Fl/2
49
3
3
1( )
2 2 2E
Fl Flw
EA EI
• Deflection at section E via superposition:
3
1 2 3
3
1 1 1
48 12 4 4
17
48 4
E E E E
Fl Flw w w w
EI EA
Fl Fl
EI EA
50
More Examples
51
More Examples
52
More Examples100 kN
P
53
ax
axaxaxxf
n
n
n0
)()(
a x
(a)
n = 0
f0 (x)
a x
(b)
n = 1
f1 (x)
a
(c)
x
n = 2
f2 (x)
Singular / Discontinuity Functions
54
01
1 1
nCaxn
dxax nn
1
00
1 naxn
nax
dx
dn
n
Calculus of Singular Functions
55
ax
axaxaxxf
n
n
n0
)()(
x
qF
A BC D E
Me
a1a2
a3l
FA FB
w 0 0 1
S 2 3
0 1
2 3
0A
A
F F x F x a q x a
F F x a q x a
1 0 1 2
1 2 3
0 1 2
1 2 3
02
2
A e
A e
qM F x M x a F x a x a
qF x M x a F x a x a
Equations of Shearing Forces & Bending Moments
56
• Denote the shearing force and bending moment at the left boundary
as FS0 and M0
0 1
S0 2 3( )sF x F F x a q x a
0 1 2
0 S0 1 2 3( )2
e
qM x M F x M x a F x a x a
• Generalized equation of bending moment
• Generalized equation of shearing forces
Boundary Conditions
57
)(xMwEI
2 1 2 3S00 1 2 3 1
2 2 6e
F F qEI M x x M x a x a x a C
2 3 2 3 40 S01 2 3 1 2
2 6 2 6 24
eM F M F qEIw x x x a x a x a C x C
0 1 2
0 S0 1 2 3( )2
e
qM x M F x M x a F x a x a
1 0 2 0,C EI C EIw
Deflection and Slope by Singular Functions
58
• FS0, M0, 0 and w0 denote the boundary values of shearing force,
bending moment, deflection and slope
Ax
MA
(a) Fixed support
w
FA
FS0 = FA, M0 = MA
0 = 0, w0 = 0FS0 = FA, M0 = 0
0 0, w0 = 0FS0 = 0, M0 = 0
0 0, w0 0
A
FA
(b) Hinged support
w
x A
w
(c) Free end
Boundary Values
59
• Find the deflection at section C and the slopes at sections A and B for
the simply supported beam shown.q
w
xA B
ll/2
C
• Solution
2 3 3S00 0
2! 3! 3! 2
F q q lEI EI M x x x x
2 3 4 40 S00 0
2! 3! 4! 4! 2
M F q q lEIw EIw EI x x x x x
Sample Problem
q
1. Equations of deflection and slope
60
• Determine boundary values
S0 0 0
3, 0 , 0
8AF F ql M w
• Determine 0 from the boundary condition at the movable hinged
support B:
3 4 34
0 0
3 30 0
8 6 24 24 16 128x l
ql l q q l qlEIw EI l l
EI
3 23 3
3 34 4
3 3
128 8 2 6 6 23 3
128 8 6 24 24 2
ql ql x q q lEI x x
ql x ql x q q lEIw x x
61
3323
26628
3
128
3
lx
qx
qxqlqlEI
4433
2242468
3
128
3
lx
qx
qxqlxqlEIw
2. The Slopes A and B and the deflection wC
EI
qlA
128
3 3
0
EI
ql
EI
qllxB
384
7)
86
1
6
1
16
3
128
3(
33
EI
ql
EI
qlww lxC
768
5
1624
1
848
3
2128
3 44
2
)(
62
• Geometric properties of the elastic
curve can be used to determine
deflection and slope.
• Consider a beam subjected to arbitrary
loading,
Moment-Area Theorems
2
2
D D
C C
D
C
x
x
x
D C
x
d d y M Md dx
dx dx EI EI
Mdx
EI
• First Moment-Area Theorem:
area under (M/EI) diagram
between C and D.
D C
63
• Second Moment-Area Theorem:
The tangential deviation of C with
respect to D is equal to the first
moment with respect to a vertical axis
through C of the area under the (M/EI)
diagram between C and D.
• Tangents to the elastic curve at P and P’
intercept a segment of length dt on the
vertical through C.
= tangential deviation of
C with respect to D
1 1
1
D
C
x
C D
x
Mdt x d x dx
EI
Mt x dx
EI
64
Moment-Area Theorems
• Cantilever beam - Select tangent at A
as the reference.
• Simply supported, symmetrically
loaded beam - select tangent at C as
the reference.
Application to Cantilevers & Beams under Symmetric Loading
0, 0A A
D D A
D D A
θ y
y t
max0, C C
D D C
B C C B C
D C D C
θ y y
y y y t
y y t
65
• Determination of the change of slope and
the tangential deviation is simplified if the
effect of each load is evaluated separately.
• Construct a separate (M/EI) diagram for
each load.
- The change of slope, θD/C, is obtained
by adding the areas under the diagrams.
- The tangential deviation, tD/C is
obtained by adding the first moments
of the areas with respect to a vertical
axis through D.
• Bending moment diagram constructed
from individual loads is said to be drawn
by parts.
Bending Moment Diagrams by Parts
66
For the prismatic beam shown,
determine the deflection and slope
at E.
SOLUTION:
• Determine the reactions at
supports.
• Construct shear, bending
moment and (M/EI) diagrams.
• Taking the tangent at C as the
reference, evaluate the slope
and tangential deviations at E.
Sample Problem
67
SOLUTION:
• Determine the reactions at
supports.
waRR DB
• Construct shear, bending
moment and (M/EI) diagrams.
EI
waa
EI
waA
EI
LwaL
EI
waA
623
1
422
32
2
22
1
68
• Slope at E:
EI
wa
EI
LwaAA
CECECE
64
32
21
aLEI
waE 23
12
2
1 2 1
3 2 2 4 2 2
3
4 4 4
4 16 8 16
E E D E C D Cy t t t
L a LA a A A
wa L wa L wa wa L
EI EI EI EI
aLEI
wayE 2
8
3
• Deflection at E:
69
• Define reference tangent at support A.
Evaluate θA by determining the tangential
deviation at B with respect to A.
• The slope at other points is found with
respect to reference tangent.
ADAD
• The deflection at D is found from the
tangential deviation at D.
B A
B A
D B A D A
FE x xFE t
t L L
xy FD FE DE t t
L
Application to Beams under Unsymmetric Loadings
B A
A
t
L
70
• Maximum deflection occurs at
point K where the tangent is
horizontal.
• Point K may be determined by
measuring an area under the (M/EI)
diagram equal to -θA .
• Obtain wmax by computing the first
moment with respect to the
vertical axis through A of the area
between A and K.
Maximum Deflection
0
B A
A
K A K A
K A A
t
L
71
Wmax: Maximum deflection
θmax: Maximum slope
[w], [θ]: Maximum allowable deflection and slope
• Stiffness calculation include:
- Stiffness check
- Rational design of cross-sections
- Find the maximum allowable loads
• θmax ≤ [θ]• wmax ≤ [w]
Stiffness Condition
72
• Deformation of beams under bending is influenced by not only
beam supports and loading condition, but also beam material,
cross-section size and shape, and beam span.
Ways to Increase Flexural Rigidity
- Increase EI
73
- Decrease beam span / increase supports
- Improve loading
- Rational design of cross-sections
• Given: l = 8 m, Iz = 2370 cm4, Wz = 237 cm3, [w] = l/500, E =
200 Gpa, [σ] = 100 Mpa.
• Find: 1. the maximum allowable load from the stiffness
condition; 2. Strength check.
Sample Problem
P20I a
z
2l2l
74
• Solution
• The strength condition is satisfied.
3
max
2
[ ]48 50048
7.11kN500
[ ] 7.11 kN
Pl lw w
EIEI
Pl
P
maxmax 60 MPa [ ]
4z z
M Pl
W W
75
d
M
2 2
0
1,
1 ( )
2 2 2
L
M Mdxdx d dθ
EI EI
M dx M xdU Md U dx
EI EI
Bending Strain Energy
76
Neutral
Surface
1
2
2
ρ
dθ
o1 o2
a b
2 2 2
2
2 22
2
0 0
2 2
2 2
L L
A
M yU dV dV
E EI
M My dA dx dx
EI EI
• Strain energy density:2 21
2 2 2
Eu
E
• Total strain energy calculated from density
• Total strain energy calculated from work
done by bending moment w.r.t. rotation
• The Elastic Curve, Deflection & Slope （挠曲线、挠度和转角）
• Differential Equation of the Elastic Curve（挠曲线微分方程）
• Deflection & Slope by Integration（积分法求挠度和转角）
• Boundary Conditions（边界条件）
• Symmetry Conditions（对称性条件）
• Continuity Conditions（连续性条件）
• Direct Integration from Distributed Loads（直接由分布荷载积分求挠度和转角）
• Direct Integration from Transverse Loads（直接由剪力积分求挠度和转角）
• Deformations in a Transverse Cross Section（梁横截面内的变形）
• Curvature Shortening（梁由于弯曲造成的轴向位移）
Contents
77
• Deflection & Slope by Superposition（叠加法求挠度和转角）
• Superposition of Loads（荷载叠加法）
• Superposition of Rigidized Structures（刚化叠加法）
• Combined Superposition（荷载和变形组合叠加法）
• Deflection & Slope by Singular Functions（奇异函数法求挠度和转角）
• Deflection & Slope by Moment-Area Theorems（图乘法求挠度和转角）
• Stiffness Condition（刚度条件）
• Ways to Increase Flexural Rigidity（梁的刚度优化设计）
• Bending Strain Energy（弯曲应变能）
Contents
78