Chapter 10
Statically Indeterminate Beams
10.1 Introduction
in this chapter we will analyze the beam in which the number of
reactions exceed the number of independent equations of equilibrium
integration of the differential equation, method of superposition
compatibility equation (consistence of deformation)
10.2 Types of Statically Indeterminate Beams
the number of reactions in excess of the number of equilibrium equations
is called the degree of static indeterminacy
1
the excess reactions are called static redundants
the structure that remains when the redundants are released is called
released structure or the primary structure
10.3 Analysis by the Differential Equations of the Deflection Curve
EIv"
=
M
EIv'"
=
V
EIviv =
-q
the procedure is essentially the same as that for a statically determine
beam and consists of writing the differential equation, integrating to obtain
its general solution, and then applying boundary and other conditions to
evaluate the unknown quantities, the unknowns consist of the redundant
reactions as well as the constants of integration
this method have the computational difficulties that arise when a large
number of constants to be evaluated, it is practical only for relatively simple
case
Example 10-1
a propped cantilever beam
AB
supports a uniform load q
determine the reactions, shear forces,
bending moments, slopes, and deflections
choose RB
as the redundant, then
2
RA
=
qL
-
RB
qL2
= CC
2
MA
-
RBL
and the bending moment of the beam is
M
=
=
RAx
qLx
-
-
qx2
CC
2
MA
-
RBx
qL2
- CC
2
EIv" = M =
qLx
-
-
RBL
qx2
- CC
2
qL2
- CC
2
RBx
-
RBL
qx2
- CC
2
-
qx3
CC
6
+
EIv' =
RBx2
qL2x
qLx2
CC - CC - CC
2
2
2
EIv =
qx4
qLx3 Rbx3 qL2x2 RBLx2
CC - CC - CC - CCC - CC + C1x + C2
6
6
4
2
24
-
RBLx
boundary conditions
v(0) =
0
v'(0)
=
0
v(L)
it is obtained
C1 = C2 = 0
and
RA
=
5qL/8
MA
=
qL2/8
RB = 3qL/8
the shear force and bending moment are
V
=
RA
- qx
=
5qL
CC
8
3
- qx
=
0
C1
M
=
RAx
-
MA
-
=
qL2
5qLx
CC - CC
8
8
qx2
CC
2
-
qx2
CC
2
the maximum shear force is
Vmax
=
5qL/8
at the fixed end
the maximum positive and negative moments are
Mpos
=
9qL2/128
-qL2/8
Mneg =
slope and deflection of the beam
v' =
qx
CC (-6L2 +
48EI
15Lx
- 8x2)
v =
qx2
- CC (3L2 48EI
5Lx +
2x2)
to determine the
-6L2 +
we have x1 =
max
=
max,
15Lx
set v' =
- 8x2
=
0
0
0.5785L
- v(x1) =
qL4
0.005416 CC
EI
the point of inflection is located at
M =
0,
i.e. x
<
0 and
M
<
0
for x
<
L /4
>
0 and
M
>
0
for x
>
L /4
4
= L /4
the slope at
B
B is
=
qL3
CC
48EI
(y')x=L =
Example 10-2
a fixed-end beam
concentrated load
P
ABC
supports a
at the midpoint
determine the reactions, shear forces,
bending moments, slopes, and deflections
because the load P
in vertical direction and symmetric
HA
= HB =
MA
=
MB
M
=
Px
C
2
EIv"
=
0
RA
=
RB
= P/2
(1 degree of indeterminacy)
-
M
MA (0 ≦ x ≦ L/2)
Px
C 2
=
MA
(0 ≦ x ≦ L/2)
after integration, it is obtained
Px2
EIv' = CC
4
EIv
Px3
= CC
12
- MA x
-
+ C1
MAx2
CC +
2
boundary conditions
v(0) =
0
v'(0)
=
0
symmetric condition
5
C1x
(0 ≦ x ≦ L/2)
+
C2
(0 ≦ x ≦ L/2)
v'(0)
=
the constants
moment
C1,
C2
and the
are obtained
MA
C1
0
=
C2 = 0
PL
= CC = MB
8
MA
the shear force and bending moment
diagrams can be plotted
thus the slope and deflection equations are
v' =
Px
- CC (L
8EI
-
2x)
(0 ≦ x ≦ L/2)
v =
Px2
- CC (3L
48EI
- 4x)
(0 ≦ x ≦ L/2)
the maximum deflection occurs at the center
max
=
- v(L/2)
=
PL3
CCC
192EI
the point of inflection occurs at the point where
L/4, the deflection at this point is
=
- v(L/4)
=
PL3
CCC
384EI
which is equal max/2
6
M =
0,
i.e.
x
=
10.4 Method of Superposition
1. selecting the reaction redundants
2. establish the force-displacement relations
3. consistence of deformation (compatibility equation)
consider a propped cantilever beam
(i) select RB
as the redundant, then
RA = qL - RB
qL2
MA = CC - RBL
2
force-displacement relation
qL4
(B)1 = CC
8EI
RBL3
(B)2 = CC
3EI
compatibility equation
B
=
qL4
CC
8EI
RB
(B)1
=
- (B)1 =
0
RBL3
CC
3EI
3qL
= CC
8
=>
(ii) select the moment MA
qL MA
RA = C + C
2
L
RA
5qL
= CC
8
as the redundant
qL MA
RB = C - C
2 L
7
MA
qL2
= CC
8
force-displacement relation
qL3
= CC (A)2
24EI
(A)1
MAL
= CC
3EI
compatibility equation
A
=
(A)1 -
thus
MA
=
qL2/8
and
RA
=
5qL/8
qL3
CC 24EI
(A)2 =
RB
=
3qL/8
Example 10-3
a continuous beam
uniform load
ABC
supports a
q
determine the reactions
select RB
RA
=
as the redundant, then
RC
=
qL -
qL
C
2
force-displacement relation
(B)1
5qL(2L)4
= CCCC
384EI
(B)2
RB(2L)3
= CCC =
48EI
=
5qL4
CC
24EI
RBL3
CC
6EI
compatibility equation
8
MAL
CC
3EI
=
0
B
=
(B)1
- (B)2 =
thus
RB
=
5qL/4
and
RA
=
RC
=
5qL4
RBL3
CC - CC
24EI
6EI
=
0
3qL/8
Example 10-4
a fixed-end beam AB is loaded by a
force
P
acting at point D
determine reactions at the ends
also determine D
this is a 2-degree of indeterminacy problem
select MA
and MB
Pb
RA = C
L
Pa
RB = C
L
+
-
as the redundants
MA
MB
C - C
L
L
MA
MB
C + C
L
L
force-displacement relations
(A)1
(A)2
(A)3
Pab(L + b)
= CCCCC
(B)1
6LEI
MAL
MAL
= CC (B)2 = CC
3EI
6EI
MBL
MBL
= CC (B)3 = CC
6EI
3EI
compatibility equations
9
Pab(L + a)
= CCCCC
6LEI
i.e.
A
=
(A)1 -
(A)2 - (A)3 =
0
B
=
(B)1 -
(B)2 - (B)3 =
0
MAL
CC +
3EI
MAL
CC +
6EI
MBL
CC
6EI
MBL
CC
3EI
Pab(L + b)
CCCCC
6LEI
Pab(L + a)
CCCCC
6LEI
=
=
solving these equations, we obtain
Pab2
= CC
L2
MA
MB
Pa2b
= CC
L2
and the reactions are
RA
Pb2
= CC (L + 2a)
L3
D can be expressed as
the deflection
D
=
RB
Pa2
= CC (L + 2b)
L3
(D)1
-
(D)2
(D)1 =
Pa2b2
CCC
3LEI
(D)2 =
MAab
CCC (L
6LEI
(D)3 =
MBab
CCC (L
6LEI
thus
D
=
Pa3b3
CCC
3L3EI
if
a
= b
-
+
+
(D)3
b)
a)
= L/2
10
=
Pa2b3
CCC (L
6L3EI
+
b)
=
Pa3b2
CCC (L
6L3EI
+
a)
then
C
and
PL
= CC
8
=
MB
=
PL3
CCC
192EI
MA
RA
=
RB
=
P
C
2
Example 10-5
a fixed-end beam
uniform load
q
AB
supports a
acting over part of the
span
determine the reactions of the beam
to obtain the moments caused by
replace
to
P
to
qdx,
a
to
dMA
qx(L - x)2dx
= CCCCC
L2
dMB
qx2(L - x)dx
= CCCCC
L2
x,
qdx,
and
b
L-x
integrating over the loaded part
MA
MB
= ∫dMA
q a
= C∫ x(L - x)2dx =
L2 0
qa2
CC (6L2 - 8aL + 3a2)
12L2
= ∫dMB
q a
= C∫ x2(L - x)dx
L2 0
qa3
CC (4L2 - 3a)
12L2
Similarly
11
=
dRA
q(L - x)2(L + 2x)dx
= CCCCCCCC
L3
dRB
qx2(3L - 2x)dx
= CCCCCC
L3
integrating over the loaded part
q a
qa
2
RA =∫dRA = C∫ (L - x) (L + 2x)dx = CC (2L3 - 2a2L + a3)
L3 0
2L3
q a
= C∫ x2(3L - 2x)dx =
L3 0
RB = ∫dRB
for the uniform acting over the entire length, i.e.
MA
RA
=
=
MB
qL2
= CC
12
RB
qL
= C
2
qa3
CC (2L - a)
2L3
a
=
L
the center point deflections due to uniform load and the end moments are
(C)1
C
=
5qL4
= CCC
384EI
MAL
(C)2 = CC =
8EI
(C)1
qL4
= CCC
384EI
-
(C)2
Example 10-6
a beam
and
ABC
rests on supports
A
B and is supported by a cable at C
12
(qL2/12)L2
qL4
CCCC = CC
8EI
96EI
find the force T
of the cable
take the cable force
the deflection
T
(C)1
as redundant
due the uniform
load can be found from example 9.9 with
a=L
qL4
= CCC
4EbIb
(C)1
the deflection (C)2 due to a force T
acting on
C
is obtained
use conjugate beam method
(C)2
= M
=
TL2
CCC L
3EbIb
=
+
2TL3
CCC
3EbIb
the elongation of the cable is
Th
= CC
EcAc
(C)3
compatibility equation
(C)1
=
(C)3
qL4
2TL3
CC - CC =
3EbIb
4EbIb
Th
CC
EcAc
T
- (C)2
=
3qL4EcAc
CCCCCCCC
8L3EcAc + 12hEbIb
13
TL L 2L
CC C C
EbIb 2 3
10.5 Temperature Effects
10.6 Longitudinal Displacements at the Ends of the Beams
14