Chapter 9
Deflections of Beams
9.1 Introduction
in this chapter, we describe methods for determining the equation of the
deflection curve of beams and finding deflection and slope at specific points
along the axis of the beam
9.2 Differential Equations of the Deflection Curve
consider a cantilever beam with a
concentrated load acting upward at the free
end
the deflection
in the
y
v
is the displacement
direction
the angle of rotation
of the axis
(also called slope) is the angle between the
x
axis and the tangent to the deflection
curve
point m1 is located at distance x
point m2 is located at distance x + dx
slope at
m1
is
slope at
m2
is
denote O'
+d
the center of curvature and
the radius of curvature, then
d
=
ds
and the curvature
is
1
=
1
C =
d
C
ds
the sign convention is pictured in figure
slope of the deflection curve
dv
C
dx
for
=
tan
or
ds j dx
small
=
dv
tan-1 C
dx
cos j 1
=
1
C =
d
C
dx
and
=
1
C =
d
C =
dx
d 2v
CC
dx2
tan
=
j ,
then
dv
C
dx
if the materials of the beam is linear elastic
=
1
C =
M
C
EI
[chapter 5]
then the differential equation of the deflection curve is obtained
d
C
dx
d2v
= CC
dx2
=
M
C
EI
it can be integrated to find
∵
dM
CC
dx
=
then
d 3v
CC
dx3
V
= C
EI
V
and v
dV
CC
dx
=
-q
d 4v
CC
dx4
=
q
-C
EI
2
sign conventions for M,
V
and
q are shown
the above equations can be written in a simple form
EIv"
=
M
EIv"' =
V
EIv""
=
-q
this equations are valid only when Hooke's law applies and when the
slope and the deflection are very small
for nonprismatic beam [I = I(x)], the equations are
d 2v
EIx CC
dx2
=
M
d
d 2v
C (EIx CC)
dx
dx2
dM
CC
dx
=
d 2v
d2
CC (EIx CC)
dx2
dx2
=
=
dV
CC
dx
V
=
-q
the exact expression for curvature can be derived
=
1
C =
v"
CCCCC
[1 + (v')2]3/2
9.3 Deflections by Integration of the Bending-Moment Equation
substitute the expression of M(x)
into
the deflection equation then integrating to
satisfy
(i) boundary conditions
(ii) continuity conditions
(iii) symmetry conditions
to obtain the slope
and the
3
deflection
v of the beam
this method is called method of
successive integration
Example 9-1
determine the deflection of beam
AB
supporting a uniform load of intensity
also determine
and
max
q
A,
flexural rigidity of the beam is
B
EI
bending moment in the beam is
M
=
qLx
CC
2
-
q x2
CC
2
differential equation of the deflection curve
EI v" =
qLx
CC
2
-
q x2
CC
2
EI v' =
qLx2
CC 4
q x3
CC
6
Then
∵ the beam is symmetry,
0
=
qL(L/2)2
CCCC
4
+
∴
-
C1
= v' =
0
at
q (L/2) 3
CCCC
6
+
C1
4
x=L/2
then
C1
q L3 / 24
=
the equation of slope is
q
- CCC (L3
24 EI
v' =
6 L x2
-
+
4 x 3)
integrating again, it is obtained
q
- CCC (L3 x
24 EI
-
2 L x 3 + x 4)
boundary condition : v =
0
at
v =
thus we have
C2
=
x
+
C2
= 0
0
then the equation of deflection is
q
- CCC (L3 x
24 EI
v =
maximum deflection
=
max
max
L
- v(C)
2
=
-
2 L x 3 + x 4)
occurs at center
5 q L4
CCC
384 EI
(x =
L/2)
(↓)
the maximum angle of rotation occurs at the supports of the beam
A
and
=
B
v'(0)
=
=
v'(L)
q L3
- CCC
24 EI
( )
q L3
= CCC
24 EI
5
( )
Example 9-2
determine the equation of deflection
curve for a cantilever beam AB
subjected
to a uniform load of intensity q
also determine
B
and
B
flexural rigidity of the beam is
at the free end
EI
bending moment in the beam
M
q L2
- CC
2
=
+
qLx
EIv" =
q L2
- CC
2
EIy' =
qL2x
- CC +
2
boundary condition
C1
v' =
=
qx
- CC
6EI
q x2
- CC
2
+
-
q x2
CC
2
qLx2
CC 2
q x3
CC
6
qLx
v' =
=
0
+
C1
at x =
0
0
(3 L2 -
3Lx
+
x 2)
integrating again to obtain the deflection curve
v =
qx2
- CC
24EI
boundary condition
C2
=
(6 L2 -
v =
4Lx
0
+
x 2)
at x =
0
6
+ C2
0
then
qx2
- CC
24EI
v =
=
max
=
max
=
B
-
(6 L2 -
B
=
v'(L)
4Lx
+
q L3
- CC ( )
6 EI
=
- v(L) =
q L4
CC (↓)
8 EI
Example 9-4
determine the equation of deflection
curve,
A,
B,
max
and
flexural rigidity of the beam is
C
EI
bending moments of the beam
M
Pbx
CC
L
=
(0 ≦ x ≦ a)
Pbx
M = CC - P (x - a) (a ≦ x ≦ L)
L
differential equations of the deflection curve
EIv"
=
x 2)
Pbx
CC (0 ≦ x ≦ a)
L
Pbx
EIv" = CC - P (x - a) (a ≦ x ≦ L)
L
integrating to obtain
7
Pbx2
EIv' = CC +
2L
C1 (0 ≦ x ≦ a)
Pbx2
P(x - a)2
EIv' = CC - CCCC
2L
2
+
C2
(a ≦ x ≦ L)
2nd integration to obtain
EIv
Pbx3
= CC +
6L
EIv
Pbx3 P(x - a)3
= CC - CCCC + C2 x
6L
6
C1 x +
(0 ≦ x ≦ a)
C3
(a ≦ x ≦ L)
+ C4
boundary conditions
(i)
v(0)
= 0
(ii)
y(L) =
0
continuity conditions
(iii)
(i)
(ii)
(iii)
v'(a-)
v(0)
=
=
v(L)
v'(a+)
0
=
0
-
C3 =
=>
PbL3 Pb3
CC - CC + C2 L + C4
6
6
+
=
v'(a )
C1
=
C2
(iv) v(a ) =
C3
=
+
v(a )
v(a+)
=>
v'(a )
-
v(a-) =
(iv)
0
=
=>
Pba2
CC
2L
=>
Pba3
Pba3
CC + C1a + C3 = CC + C2a +
6L
6L
C4
8
+
C1
=
Pba2
CC
2L
0
+
C2
C4
then we have
C1
=
C2
=
Pb (L2 - b2)
- CCCCC
6L
C3
=
C4
=
0
thus the equations of slope and deflection are
v' =
Pb
- CC (L2 - b2 - 3x2)
6LEI
(0 ≦ x ≦ a)
v' =
Pb
- CC (L2 - b2 - 3x2) 6LEI
P(x - a )2
CCCC
2EI
v =
Pbx
- CC (L2 - b2 - x2)
6LEI
(0 ≦ x ≦ a)
v =
Pbx
- CC (L2 - b2 - x2)
6LEI
-
(a ≦ x ≦ L)
P(x - a )3
CCCC
6EI
(a ≦ x ≦ L)
angles of rotation at supports
∵
Pab(L + b)
- CCCCC
6LEI
A
=
v'(0)
=
B
=
v'(L)
Pab(L + a)
= CCCCC
6LEI
A
is function of a (or b), to find
=
A
d
A
( )
( )
( A)max,
set
d
A
/ db = 0
=
L/ 3
Pb(L2 - b2)
- CCCCC
6LEI
/ db =
0
=> L2 - 3b2
9
=
0
=> b
PL2 3
- CCCC
27 EI
( A)max =
for maximum
dv
C
dx
=
at
0
=
max
x =
occurs at x1,
=>
x1 =
if a
>
L2 - b2
CCC
3
Pb(L2 - b2)3/2
- v(x1) = CCCCC
9 3 LEI
L/2
C
=
- v(L/2)
b,
x1
<
(a
≧ b)
a
(↓)
Pb(3L2 - 4b2)
CCCCCC
48 EI
=
(↓)
∵ the maximum deflection always occurs near the midpoint, ∴
gives a good approximation of the
max
in most case, the error is less than 3%
an important special case is
v'
a
= b
= L/2
v' =
P
CC (L2 - 4x2)
16EI
v =
P
CC (3L2 - 4x2) (0 ≦ x ≦ L/2)
48EI
and v
A
max
(0 ≦ x ≦ L/2)
are symmetric with respect to
=
PL2
= CC
16EI
B
=
C
=
PL3
CC
48EI
10
x
=
L/2
C
9.4 Deflections by Integration of Shear-Force and Load Equations
the procedure is similar to that for the bending moment equation except
that more integrations are required
if we begin from the load equation, which is of fourth order, four
integrations are needed
Example 9-4
determine the equation of deflection curve
for the cantilever beam
AB
supporting a
triangularly distributed load of maximum
intensity q0
also determine
B
and
B
flexural rigidity of the beam is
q
=
EIv""
EI
q0 (L - x)
CCCC
L
=
-q
=
q0 (L - x)
- CCCC
L
the first integration gives
EIv"'
∵
thus
=
v"'(L)
EIv"'
q0 (L - x)2
- CCCC +
2L
= V
=
=
0
C1
=> C1
q0 (L - x)2
- CCCC
2L
11
=
0
2nd integration
EIv"
∵
=
v"(L)
thus
q0 (L - x)3
- CCCC +
6L
=
EIv"
M
=
=
0
C2
=>
C2
=
0
q0 (L - x)3
- CCCC
6L
3rd and 4th integrations to obtain the slope and deflection
q0 (L - x)4
EIv' = - CCCC +
24L
q0 (L - x)5
- CCCC +
120L
EIv =
boundary conditions : v'(0)
the constants
C3
C3
q 0 L3
- CC
24
=
and
C3
C3 x +
= v(0) =
C4
0
C4 can be obtained
C4
q0L4
CC
120
=
then the slope and deflection of the beam are
v' =
q 0x
- CCC (4L3
24LEI
v =
q 0x 2
- CCC (10L3 120LEI
B
=
-
6L2x
+
10L2x +
q0L3
v'(L) = - CCC
24 EI
( )
12
4Lx2
- x 3)
5Lx2 - x3)
B
=
q 0 L4
- v(L) = CCC
30 EI
(↓ )
Example 9-5
an overhanging beam
concentrated load
P
ABC
with a
applied at the end
determine the equation of deflection
curve and the deflection
at the end
C
flexural rigidity of the beam is
EI
the shear forces in parts AB and BC are
V =
P
-C
2
(0 < x < L)
V =
P
3L
(L < x < C)
2
the third order differential equations are
EIv'"
=
P
- C (0 < x < L)
2
EIv'"
=
P
3L
(L < x < C)
2
bending moment in the beam can be obtained by integration
M
=
EIv"
=
M
=
EIv" =
Px
-C +
2
Px
+
C1
C2
13
(0 ≦ x ≦ L)
3L
(L ≦ x ≦ C)
2
boundary conditions : v"(0) = v"(3L/2)
we get
C1
=
0
C2
=
= 0
3PL
- CC
2
therefore the bending moment equations are
M
=
EIv"
=
Px
-C
2
M
=
EIv"
=
P(3L - 2x)
- CCCCC
2
(0 ≦ x ≦ L)
3L
(L ≦ x ≦ C)
2
2nd integration to obtain the slope of the beam
Px2
EIv' = - CC
4
+
Px(3L - x)
EIv' = - CCCCC
2
continuity condition :
PL2
- CC
4
then
C4
+
=
C3
=
C3 +
(0 ≦ x ≦ L)
C3
3L
C4 (L ≦ x ≦ C)
2
+
v'(L-) =
v'(L+)
- PL2 +
C4
3PL2
CC
4
the 3rd integration gives
EIv =
Px3
- CC
12
EIv =
Px2(9L - 2x)
- CCCCC +
12
+
C3 x
+
C4 x
14
(0 ≦ x ≦ L)
C5
+
C6
3L
(L ≦ x ≦ C)
2
v(0) = v(L-) = 0
boundary conditions :
we obtain
C5
=
and then
0
C3
C4
5PL2
CC
6
=
=
PL2
CC
12
the last boundary condition : v(L+)
then
C6
=
=
0
PL3
- CC
4
the deflection equations are obtained
v =
Px
CC (L2
12EI
-
x 2)
(0 ≦ x ≦ L)
P
- CC (3L3 - 10L2x + 9Lx2
12EI
P
= - CC (3L - x) (L - x) (L - 2x)
12EI
v =
deflection at
C
=
-2x3)
3L
(L ≦ x ≦ C)
2
C is
3L
- v(C) =
2
PL3
CC
8EI
(↓)
9.5 Method of Superposition
the slope and deflection of beam caused by several different loads acting
simultaneously can be found by superimposing the slopes and deflections
caused by the loads acting separately
15
consider a simply beam supports two
loads : (1) uniform load of intensity
and
q
(2) a concentrated load P
the slope and deflection due to load
1
are
5qL4
= CCC
384EI
( C)1
( A )1
= ( B)1
=
qL3
CC
24EI
the slope and deflection due to load
2
are
PL3
= CC
48EI
( C)2
( A )2
= ( B)2
PL2
= CC
16EI
therefore the deflection and slope due to the combined loading are
C
A
=
=
( C)1
B
+
=
5qL4
= CCC
384EI
( C)2
( A)1
+
( A)2
=
+
PL3
CC
48EI
qL3
PL2
CC + CC
24EI
16EI
tables of beam deflection for simply and cantilever beams are given in
Appendix
G
superposition method may also be used for distributed loading
consider a simple beam ACB
with a
triangular load acting on the left-hand half
16
the deflection of midpoint due to a
concentrated load is obtained [table G-2]
C
substitute
d
Pa
CC (3L2 - 4a2)
48EI
=
q dx
for P and x for a
(qdx) x
= CCC (3L2 - 4x2)
48EI
C
the intensity of the distributed load is
q
then
d
thus
=
2q0x
CC
L
C
due to the concentrated load q dx
C
q0 x 2
= CCC (3L2 - 4x2) dx
24LEI
C
due to the entire triangular load is
q0 x2
= ∫ CCC (3L2 - 4x2) dx
0
24LEI
L/2
C
similarly, the slope
A
due to
=
acting on
x is
q 0 L4
CCC
240EI
P acting on a distance a
from left
a with
(L - x)
end is
d
A
replacing
Pab(L + b)
= CCCCC
6LEI
P
with
2q0x dx/L,
17
x,
and b with
2q0x2(L - x)(L + L - x)
q0
d A = CCCCCCCCC dx = CCC (L - x) (2L - x) x2 dx
6L2EI
3L2EI
thus the slope at A
L/2
∫
=
A
0
throughout the region of the load is
q0
CCC (L - x) (2L - x) x2 dx =
3L2EI
41q0L3
CCC
2880EI
the principle of superposition is valid under the following conditions
(1) Hooke's law holds for the material
(2) the deflections and rotations are small
(3) the presence of the deflection does not alter the actions of applied
loads
these requirements ensure that the differential equations of the deflection
curve are linear
Example 9-6
a cantilever beam
uniform load
q
AB
supports a
and a concentrated load
P
as shown
determine
EI
=
B
and
B
constant
from Appendix G :
for uniform load
( B)1
q
qa3
= CC (4L - a)
24EI
( B )1
for the concentrated load P
18
qa3
= CC
6EI
PL3
= CC
3EI
( B)2
PL2
= CC
2EI
( B )2
then the deflection and slope due to combined loading are
=
B
=
B
( B)1
+
( B )1 +
qa3
PL3
CC (4L - a) + CC
24EI
3EI
( B)2
=
( B)2
qa3
= CC
6EI
+
PL2
CC
2EI
Example 9-7
a cantilever beam
load
q
with uniform
acting on the right-half
determine
EI
AB
=
and
B
B
constant
consider an element of load has magnitude
q dx
and is located at distance x
from the
support
from Appendix G, table G-1, case 5
by replacing
d
B
=
P
with
q dx
(qdx)(x2)(3L-x)
CCCCCC
6EI
and
d
a with x
B
(qdx)(x2)
= CCCC
2EI
by integrating over the loaded region
L
B
= ∫
L/2
qx2(3L-x)
CCCC dx =
6EI
19
41qL4
CCC
384EI
qx2
CC dx
2EI
L
∫
=
B
L/2
7qL3
= CC
48EI
Example 9-8
a compound beam
concentrated load
load
q
P
supports a
and an uniform
as shown
determine
EI
ABC
=
and
B
A
constant
we may consider the beam as composed
of two parts : (1) simple beam AB,
and
(2) cantilever beam BC
the internal force
F = 2P/3
is obtained
for the cantilever beam BC
B
qb4
= CC
8EI
for the beam
parts : (1) angle
AB,
Fb3
CC
3EI
+
A
B
= C
a
the angle due to
G
=
P
with replacing a
2Pb3
CC
9EI
consists of two
BAB' produced by
(2) the bending of beam AB
( A )1
qb4
CC +
8EI
B,
and
by the load
P
qb4
= CCC
8aEI
+
2Pb3
CCC
9aEI
is obtained from Case 5 of
by 2a/3
and
20
b
by a/3
table G-2, Appendix
P(2a/3)(a/3)(a + a/3)
= CCCCCCCCC
6aEI
( A )2
note that in this problem,
continuous,
g
( B)L
i.e.
=
4Pa2
CCC
81EI
is continuous and
B
B
does not
( B )R
Example 9-9
an overhanging beam ABC
determine
EI
C
=
supports a uniform load
q as shown
C
constant
may be obtained due to two parts
(1) rotation of angle
B
(2) a cantilever beam subjected
to uniform load
q
firstly, we want to find
B
then
=
qL3
- CC
24EI
=
qL3
- CC +
24EI
1
=
a
B
MBL
CC
3EI
+
B
qa2L
CC =
6EI
=
qaL(4a2 - L2)
CCCCCC
24EI
bending of the overhang BC
2
=
qL(4a2 - L2)
CCCCC
24EI
produces an additional deflection
qa4
CC
8EI
21
2
therefore, the total downward deflection of C
=
C
for
a
large,
0
a
=
a > 0.4343 L,
point
C
=
C
D
=
1
+
qa4
CC
8EI
qa
CC (a + L) (3a2 + aL - L2)
24EI
=
for
+
1
qaL(4a2 - L2)
CCCCCC
24EI
is
is downward; for
a
3a2 + aL - L2
0
L( 13 - 1)
CCCCC
6
=
=
C
is upward
0.4343 L
is downward;
C
small,
a < 0.4343 L,
C
is upward
is the point of inflection, the curvature is zero because the
bending moment is zero at this point
d2y/dx2 =
at point D,
0
9.6 Moment-Area Method
the method is especially suitable when the deflection or angle of rotation
at only one point of the beam is desired
consider a segment
denote
B
and
B/A
AB of the beam
the difference between
A
B/A
=
consider points
B
m1
A
and
m2
22
d
ds
C =
=
Mdx / EI
strip of the
Mdx
CC
EI
is the area of the shaded
Mdx / EI
diagram
integrating both side between A
B M
∫ C dx
A EI
B
∫ d
=
A
B/A
=
-
B
and B
A
=
area of the M/EI diagram between A
and B
this is the First moment-area theorem
next, let us consider the vertical offset
between points
tB/A
B
and
B1 (on
the tangent of A)
∵
dt =
x1 d
Mdx
x1 CC
EI
=
integrating between A
B
∫ dt
A
i.e. tB/A
A and
and
B
B
Mdx
= ∫ x1 CC
A
EI
=
1st moment of the area of the M/EI
B, evaluated w. r. t. B
this if the Second moment-area theorem
Example 9-10
determine
cantilever beam
B
and
AB
B
of a
supporting a
23
diagram between
concentrated load
sketch the
B
PL2
- CC
2EI
1 PL
-CLC =
2 EI
=
B/A
Q1
at
M/EI diagram first
A1 =
B
P
=
=
B
-
=
A
=
B
PL3
- CC
6EI
2L
A1 C =
3
A1 x
=
- Q1
PL3
= CC
6EI
(↓)
Example 9-11
determine
B
cantilever beam
and
AB
of a
B
supporting a
uniform load of intensity
q
acting
over the right-half
sketch the
M/EI diagram first
A1 =
1 L qL2
C C (CC)
3 2 8EI
A2 =
L qL2
C (CC)
2 8EI
A3 =
1 L qL2
C C (CC)
2 2 4EI
=
=
PL2
- CC
2EI
qL3
CC
48EI
qL3
CC
16EI
qL3
= CC
16EI
24
(
)
=
B/A
B
B
=
A1 + A2 + A3
7qL3
= CC
16EI
=
tB/A
=
A 1 x1 +
=
qL3 1 3L
1 3L
1 5L
CC (C C + C C + C C)
EI 48 8
16 4
16 6
A 2 x2
+ A 3 x3
Example 9-12
a simple beam ADB supports a
concentrated load
determine
A1 =
A
P
as shown
and
D
L Pab
C (CC)
2 LEI
Pab
CC
2EI
=
L+b
Pab (L + b)
tB/A = A1 CCC = CCCCC
6EI
3
A
=
BB1
CC
L
=
to find the deflection
D
( )
=
DD1
D 2D 1
DD1
= a
Pab (L + b)
CCCCC
6EIL
at
D
D
- D 2D 1
A
=
Pa2b (L + b)
CCCCCC
6EIL
=
tD/A = A2 x2
=
A Pab a
C CC C
2 EIL 3
=
Pa3b
CCC
6EIL
25
=
41qL4
CCC
384EI
(↓)
Pa2b2
CCC
3EIL
=
D
to find the maximum deflection
x1 Pbx1
C CC
2 EIL
A3 =
=
E/A
E
=
Pbx12
CCC
2EIL
A
=
Pab (L + b)
= CCCCC
6EIL
A
then
x1 =
and
max
or
-
max
- A3 =
=
=
x1
A
x1
- A3 C =
3
=
offset of point
max
=
2 x1
A3 CC
3
EI d /dx
=
A
v =
=
- Pbx12 / 2EIL
=
Pb
3/2
CCCC (L2 - b2)
9 3 EIL
from tangent at E
Pb
3/2
CCCC (L2 - b2)
9 3 EIL
M
Integrating
=
E
[(L2 - b2) / 3]2
max
=
we set
Pbx12
CCC
2EIL
Conjugate Beam Method
EIv"
at E,
M
∫C dx
EI
M
∫∫ C dx dx
EI
26
0
beam theory
dM / dx
V
=
= V
dV / dx
-∫q dx
M
=
= -q
-∫∫ q dx dx
suppose we have a beam, called conjugate beam, whose length equal to
the actual beam, let this beam subjected to so-called "elastic load" of
intensity M/EI,
then the shear force and bending moment over a portion
of the conjugate beam, denoted by V and
V =
M
-∫C dx
EI
M
=
M,
can be obtained
M
- ∫∫ C dx dx
EI
then
(1) the slope at the given section of the beam equals the minus shear
force in the corresponding section of the conjugate beam
(2) the deflection at the given section of the beam equals the minus
bending moment in the corresponding section of the conjugate beam
i.e.
=
-V
=
-M
the support conditions between the actual beam and conjugate beam can
be found
Actual Beam
Conjugate Beam
fixed end
= 0, v = 0
V = 0,
free end
g 0,
V g 0, M g 0
fixed end
simple end
g
V g 0, M = 0
simple end
vg0
0,
v=0
M=0
free end
interior support
g 0,
v=0
V g 0, M = 0
interior hinge
interior hinge
g 0,
vg0
V g 0, M g 0
interior support
27
Example 1
1 PL
PL2
B = - VB = - C CC L = - CC
2 EI
2EI
( )
PL2 2L
PL3
B = - MB = - CC C = - CC
2EI 3
3EI
(↓)
Example 2
1 2L wL2
wL3
A = - VA = - C C CC = - CC
2 3 8EI
24EI
( )
wL3 L wL2 L L
C = - MC = - CC C - CC C C
24EI 2 8EI 2 4
1 wL2 L 3L
1 1
1
wL4
+ C CC C C = - (C - C + CC) CC
48 64 128
EI
3 8EI 2 8
=
5 wL4
CCC
384EI
(↓)
28
Example 3
A
= - VA = -RA
1 M 2L
ML
= - C C C = - CC ( )
2 EI 3
3EI
1 M L
ML
B = - VB = RB = C C C = CC
2 EI 3 6EI
( )
ML L 1 L M L
C = - MC = - (CC C - C C CC C)
6EI 2 2 2 2EI 6
1 1 ML2
= - (C - C) CC =
12 48 EI
ML2
- CC
16EI
(↓)
Example 4
B
=
- VB
ML
= CC
EI
ML L
B = - MB = CC C =
EI 2
(
ML2
CC
2EI
)
(↑)
Example 5
qL3
L qL2
B = - VB = - C CC = - CC
3 2EI
6EI
( )
qL3 3L
qL4
B = - MB = - CC C = - CC
6EI 4
8EI
(↓)
29
Example 6
1 L PL
PL2
A = - VA = - C C CC = - CC
2 2 4EI
16EI
( )
PL2 L 1 L PL L
C = - MC = - (CC C - C C CC C)
16EI 2 2 2 4EI 6
PL3 1 1
PL3
= - CC (C - C) = - CC
EI 32 96
48EI
(↓)
9.7 Nonprismatic Beam
EI
g constant
Example 9-13
a beam ABCDE is supported a
concentrated load P at midspan as
shown
IBD
=
2 IAB
=
2 IDE =
determine the deflection curve,
M
then
=
Px
C
2
A
and
C
L
(0 ≦ x ≦ C)
2
EIv" = Px / 2
E(2I)v" = Px / 2
2I
(0 ≦ x ≦ L/4)
(L/4 ≦ x ≦ L/2)
30
thus
∵
Px2
v' = CC + C1
4EI
(0 ≦ x ≦ L/4)
Px2
v' = CC + C2
8EI
(L/4 ≦ x ≦ L/2)
v' =
∴
C2
0
=
at x =
PL2
- CC
32EI
continuity condition v'(L/4)C1
=
therefore
=
= v'(L/4)+
5PL2
- CCC
128EI
v' =
P
- CCC (5L2
128EI
v' =
P
- CC (L2
32EI
the angle of rotation
A
L/2 (symmetric)
v'(0)
A
=
32x2)
-
- 4x2)
(0 ≦ x ≦ L/4)
(L/4 ≦ x ≦ L/2)
is
5PL2
- CCC
128EI
( )
integrating the slope equation and obtained
v =
P
- CCC (5L2x
128EI
32x3
- CC)
3
+
v =
P
- CC (L2x 32EI
4x3
CC)
3
C4
31
+
C3
(0 ≦ x ≦ L/4)
(L/4 ≦ x ≦ L/2)
boundary condition
v(0) =
we get
C3
=
0
0
continuity condition v(L/4)- =
we get
C4
v(L/4)+
PL3
- CCC
768EI
=
therefore the deflection equations are
v =
Px
- CCC (15L2 384EI
v =
P
- CCC (L3
768EI
+
32x2)
24L2x -
(0 ≦ x ≦ L/4)
32x3)
(L/4 ≦ x ≦ L/2)
the midpoint deflection is obtained
=
C
- v(L/2)
=
3PL3
CCC
256EI
(↓)
moment-area method and conjugate beam methods can also be used
Example 9-14
a cantilever beam
concentrated load
IBC =
2 IAB
determine
denote
1
1
ABC
supports a
P
at the free end
=
2I
A
the deflection of A due to C fixed
P(L/2)3
= CCC
3EI
=
PL3
CCC
24EI
32
and
P(L/2)3
(PL/2)(L/2)2
5PL3
CCC + CCCCC = CC
3E(2I)
2E(2I)
96EI
C
=
C
P(L/2)2 (PL/2)(L/2)
= CCC + CCCCC
2E(2I)
E(2I)
addition deflection at A due to
=
2
=
A
C
1
+
+
L
CC
2
2
C
=
=
and
PL2
= CC
16EI
C
5PL3
CC
48EI
5PL3
CC
16EI
moment-area method and conjugate beam
methods can also be used
9.8 Strain Energy of Bending
consider a simple beam
AB
subjected
to pure bending under the action of two
couples M
the angle
=
is
L
C =
L
ML
CC
EI
=
if the material is linear elastic, M
and
has linear relation, then
W =
U
=
M
CC
2
=
M 2L
EI 2
CC = CC
2EI
2L
33
for an element of the beam
d
=
dU
dx
d 2y
= CC dx
dx2
Md
CC
2
= dW =
M 2dx
CCC
2EI
=
EI(d )2
CCC
2dx
=
by integrating throughout the length of the beam
U
=
M 2dx
∫ CCC
0
2EI
L
EI d 2y 2
∫ C (CC) dx
0
2 dx2
L
=
shear force in beam may produce energy, but for the beam with
L/d > 8,
the strain energy due to shear is relatively small and may be disregarded
deflection caused by a single load
U
=
=
W =
2U
CC
P
P
C
2
U
or
=
=
W =
2U
CC
M0
Example 9-15
a simple beam
AB
of length
L
supports a uniform load of intensity q
evaluate the strain energy
M
=
qLx
CC
2
qx2
- C =
2
q
C (Lx - x2)
2
34
M0
CC
2
M2dx
∫ CC =
0
2EI
L
U
=
1
q
L
2
CC ∫ [C(Lx - x2)] dx
2EI 0 2
q2
L
CC∫ (L2x2 - 2Lx3 + x4)dx
8EI 0
=
q2L5 / 240EI
=
Example 9-16
a cantilever beam AB
is subjected to
three different loading conditions
(a) a concentrated load P at its free end
(b) a moment
(c) both
M0 at its free end
P and
M0
acting simultaneously
determine
A
due to loading (a)
determine
A
due to loading
(a)
=
M
U
2
M2dx
L (-Px) dx
= ∫ CC = ∫ CCC =
0
0
2EI
2EI
M
U
- Px
L
P A P2L3
CC = CC
2
6EI
W=U
(b)
(b)
=
W=U
PL3
A = CC
3EI
- M0
M2dx
= ∫ CC =
0
2EI
L
P 2 L3
CC
6EI
(-M0)2dx
∫ CCCC
0
2EI
L
M 02L
M0 A
CC = CC
2
2EI
35
A
=
M 02L
CC
2EI
M0L
= CC
EI
(c)
M
=
- Px
M2dx
U = ∫ CC
0
2EI
L
-
(-Px - M0)2dx
P 2 L3
PM0L2 M02L
=∫ CCCCC = CC + CCC + CC
0
2EI
6EI
2EI
2EI
L
P A
CC
2
W=U
M0
+
M0 A
CC =
2
1 equation for two unknowns
P2L3
CC
6EI
and
A
+
PM0L2
CCC
2EI
+
M02L
CC
2EI
A
9.9 Castigliano's Theorem (Energy Method)
dU
= Pd
dC =
where
C
dP
dU
CC
d
=
dC
CC
dP
=
is complementary strain energy
for linear elastic materials C
then we have
dU
CC
dP
=
similarly
dU
CC
dM
=
for both P
and
∂U
CC
∂P
P
M
=
U
acting simultaneously, U
∂U
=
CC
∂M
=
in example 9-16 (c)
36
=
U(P, M)
U
=
P 2 L3
CC
6EI
+
+
=
PL3
CC
6EI
+
M 0 L2
CC
2EI
=
PL2
CC
2EI
+
M 0L
CC
EI
∂U
=
CC
∂P
∂U
=
CC
M 02L
CC
2EI
PM0L2
CCC
2EI
∂M
in general relationship
∂U
=
i
CC
Castigliano's Theorem
∂P i
∂U
=
i
CC
M2dx
M ∂M
CC∫CC = ∫(C)(CC)dx
∂P i
2EI
EI ∂P i
∂
=
∂P i
this is the modified Castigliano's Theorem
in example 9-16 (c)
M
=
∂M
CC
∂P
- Px
=
-x
-
M0
∂U
CC
∂M
=
-1
0
=
1
C∫(-Px - M0)(-x)dx
EI
=
1
C ∫(-Px - M0)(-1)dx =
EI
=
37
PL3
CC +
6EI
M 0 L2
CC
2EI
PL2
CC
2EI
M 0L
CC
EI
+
Example 9-17
a simple beam
uniform load
q
concentrated load
AB
=
P
20 kN/m, and a
= 25 kN
L
=
2.5 m
I
=
31.2 x 102 cm4
determine
M
=
supports a
E =
210 GPa
C
Px
CC
2
+
qLx
CC
2
qx2
CC
2
-
method (1)
U
M2dx
= ∫CC
2EI
=
P2L3
= CC +
96EI
∂U
C
=
CC
∂P
1
Px
2∫ CC(CC
0
2EI 2
qLx
+ CC
2
L/2
5PqL4
CCC
384EI
PL3
= CC
48EI
-
qx2 2
CC) dx
2
q 2 L5
CCC
240EI
+
5qL4
CCC
384EI
+
method (2)
∂M / ∂P =
C
x/2
M ∂M
= ∫(C)(CC)dx =
EI ∂P
=
PL3
CC
48EI
+
1 Px qLx
qx2 x
2∫ C(C + CC - CC)(C) dx
0
EI 2
2
2
2
L/2
5qL4
CCC
384EI
= 1.24 mm +
1.55 mm
38
=
2.79 mm
Example 9-18
a overhanging beam
ABC
supports a
uniform load and a concentrated load as
shown
determine
and
C
the reaction at A
due to the loading is
qL
RA = C 2
P
C
2
qx12
RA x1 - CC
2
=
MAB
C
qLx1
CC
2
=
-
Px1
CC
2
qx12
CC (0 ≦ x1 ≦ L)
2
-
= - Px2 (0 ≦ x1 ≦ L/2)
MBC
then the partial derivatives are
∂ MAB / ∂ P
=
- x1/2
(0 ≦ x1 ≦ L)
∂ MBC / ∂ P
=
- x2
(0 ≦ x2 ≦ L/2)
C
=
∫(M/EI)( ∂ M / ∂ P)dx
L
= ∫ (MAB /EI)( ∂ MAB / ∂ P)dx
0
L/2
+ ∫ (MBC/EI)( ∂ MBC / ∂ P)dx
0
1 L qLx1 Px1
qx12 x1
1 L/2
= C∫ (CC - CC - CC)(- C)dx1 + C∫ (-Px2)(-x2)dx2
EI 0 2
2
2
2
EI 0
=
PL3
CC
8EI
qL4
- CC
48EI
39
to determine the angle
qL
RA = C
2
MAB
C,
P
C
2
-
=
=
qLx1
CC
2
= - Px2
MBC
MC
at C
MC
- CC
L
qx12
RA x1 - CC
2
-
we place a couple of moment
Px1
CC
2
qx12
CC
2
MC x 1
- CC L
(0 ≦ x1 ≦ L)
(0 ≦ x1 ≦ L/2)
- MC
then the partial derivatives are
∂ MAB / ∂ MC
=
- x1/L
(0 ≦ x1 ≦ L)
∂ MBC / ∂ MC
=
-1
(0 ≦ x2 ≦ L/2)
C
=
∫(M/EI)(M/ MC)dx
L
= ∫ (MAB /EI)(MAB / MC)dx
0
+
L/2
∫
(MBC /EI)(MBC / MC)dx
0
1 L qLx1
Px1
MC x1
= C∫ (CC - CC - CC EI 0
2
2
L
1 L/2
+ C∫ (-Px2 - MC)(-1)dx2
EI 0
since
MC is a virtual load,
set MC = 0,
obtained
C
7PL2
= CC 24EI
qL4
CC
24EI
40
qx12
x1
CC)(- C)dx1
2
L
after integrating
C
is
9.10 Deflections Produced by Impact
9.11 Temperature Effects
41