Post on 24-Apr-2020
Statically Indeterminate Structures
mi@seu.edu.cn
• Statically Determinate Problems(静定问题)
• Statically Indeterminate Problems(超静定问题)
• Degrees of Indeterminacy(超静定次数)
• Advantages & Disadvantages of Indeterminate Structures
(超静定结构的优缺点)
• Redundancy & Basic Determinate System(冗余约束与基本静定系)
• General Ideas for Analyzing Indeterminate Structures
(超静定问题一般解法)
Contents
2
• Force Method Solution Procedure(力法)
• Statically Indeterminate Bars(超静定拉压杆)
• Assembly Stress(装配应力)
• Thermo Stress(热应力)
• Statically Indeterminate Shafts(超静定扭转轴)
• Statically Indeterminate Beams(超静定弯曲梁)
• Moment-Area Theorems with Indeterminate Beams(图乘法求解超静定梁)
• Combined Indeterminate Structures (组合超静定结构)
Contents
3
• Problems can be completely solved via static equilibrium
• Number of unknowns (forces/moments) = number of
independent equations from static equilibration
0 ?0 ?
?0
x Ax
y Ay
ByA
F FF F
FM
Statically Determinate Problems
4
A B
A BC
• Problems cannot be solved from static equilibrium alone
• Number of unknowns (forces/moments) > number of
independent equations from static equilibration
?0 ?
0 ?
?0?
Ax
x Ay
y By
CxA
Cy
FF F
F F
FMF
Statically Indeterminate Problems
5
• For a co-planer structure, there are at most three
equilibrium equations for each portion of the structure. If
there is a total of n portions and m unknown reaction
forces from supports:
3 Statically determinate
3 Statically
- 3 Degrees of indetermi
indet
n
erminate
acy
m n
m n
m n
Statically determinate Degrees of Indeterminacy = 2
Degrees of Indeterminacy of Structures
6
• Advantages:
Advantages & Disadvantages of Indeterminate Structures
- Redistribution of reaction forces / internal forces
- Smaller deformation
- Greater stiffness as a whole structure
• Disadvantages:
- Thermal and residual stresses due to temperature change
and fabrication errors
7
• Redundancy: unnecessary restraints without which the
static equilibrium of a structure still holds.
• Basic determinate system: the same structure as of a
statically indeterminate system after replacing redundant
restraints with extra constraining loads
Redundancy & Basic Determinate System
8
• Basic determinate structure: obtained via replacing
redundant restraints with extra constraining loads.
How to Analyze Statically Indeterminate Structures?
• Equilibrium: is satisfied when the reaction forces at
supports hold the structure at rest, as the structure is
subjected to external loads
• Deformation compatibility: satisfied when the various
segments of the structure fit together without intentional
breaks or overlaps
• Deformation-load relationship: depends on the manner
the material of the structure responds to the applied loads,
which can be linear/nonlinear/viscous and elastic/inelastic;
for our study the behavior is assumed to be linearly elastic
9
• The method of consistent deformations or force method
was originally developed by James Clerk Maxwell in 1874
and later refined by Otto Mohr and Heinrich Műller-
Breslau.
Christian Otto Mohr
(1835 – 1918)
James Clerk Maxwell
(1831 – 1879)
Heinrich Műller-Breslau
(1851 - 1925)
Force Method (Method of Consistent Deformations)
10
• Making the indeterminate structure determinate
Force Method Solution Procedure
• Obtaining its equations of equilibrium
• Analyzing the deformation compatibility that must be
satisfied at selected redundant restraints
• Substituting the deformation-load relationship into
deformation compatibility, resulting in complementary
equations
• Joining of equilibrium equations and complementary
equations
• Types of Problems can be addressed include (a) bar
tension/compression; (b) shaft torsion; and (c) beam
bending11
Statically Indeterminate Bars
• For an axially loaded bar with both ends fixed, determine
the reaction forces.
12
1
2
0BB
B
A
P R LR LL
EA EA
LR P
L
LR P
L
• Solution:
12
• Determine the reaction forces at the bar ends.
6
3
3 64
31
6
3
6
400 10
600 10
150 10 400 100
600 10
250 10
900 10
250 10
577kN
900 577 323kN
B
B
i i
i i B
B
B
A
R
R
F LL
EA E R
R
R
R
• Solution:
Sample Problem
• Solution:
Sample Problem
• Determine the reaction forces, if any, at the bar ends. E = 200 Gpa.
43
1
6
3
3 6
9 3
6
3
6
4.5 10
400 10
600 10
150 10 400 10
200 10 600 10
250 10
900 10
250 10
115.4kN
900 115.4 784.6kN
i i
i i
B
B
B
B
B
A
F LL
EA
R
R
R
R
R
R
14
Sample Problem
• What is the deformation of the
rod and tube when a force P is
exerted on a rigid end plate as
shown?
• Solution:
1 2
1 21 2
1 1 2 2
1 1 2 21 2
1 1 2 2 1 1 2 2
1 1 2 2
,
P P P
PL P LL L
E A E A
E A P E A PP P
E A E A E A E A
PLL
E A E A
15
• Given: tension/compression rigidities E1A1 = E2A2 = EA,
E3A3, external load F. Find: internal forces in bars.
1. Make the truss determinate by
replacing extension/contraction
restraint along Bar 3 with FN3
F
A
1 23
l
1 2
1 2 3
0
0 ( )cos
x N N
y N N N
F F F
F F F F F
A
F
FN3FN1 FN2
Sample Problem
• Solution:
2. Static equilibrium at joint A
16
4. Displacement-force relationship
3. Displacement compatibility at joint A
Due to symmetry
21 ll
cos31 ll
11
cos
NF ll
EA
33
33
AE
lFl N
A
1 23
l
A*
17
5. Complementary equation
21 31 3 1 3
1 1 3 3 3 3
cos cos coscos
N NN N
F l F l EAl l F F
E A E A E A
6. Joint of force equilibrium and complementary equations
1 23 3
1 2 2
1 2 3
323
1 3
3 33 3
2coscos
( )cos
cos 1 2 cos
N N
N N
N N N
N
N N
FF F
E AF F
EAF F F F
FFEA EAF F
E A E A
18
• Given: tension/compression rigidities E1A1, E2A2, E3A3,
external load F. Find: internal forces in bars.
F
A
1 23
l
Sample Problem
• Solution:
1. Make the truss determinate by
replacing extension/contraction
restraint along Bar 3 with FN3
1 2
1 2 3
sin sin 1
cos cos 0 2
N N
N N N
F F F
F F F
2. Static equilibrium at joint A
FN3FN1 FN2
FA
19
3. Displacement compatibility at joint A
1l
2
A*2l
3l
32 1
2 1 3
2sin sin tan
2 cos
ll l
l l l
4. Complementary equation
1 2 31 2 3
1 1 2 2 3 3
2 1 3
21 2 3
1
2 2 1 1 3
1 2 2 3 3
3
2 cos 0
, , cos cos
2 coscos cos
3
N N N
N
N
N N
N N
F l F l F ll l l
E A E A E A
F l F l F l
E A E A E A
F F F
E A E A E A
5. Joint of force equilibrium and complementary equations
A
1 23
l
• Residual stresses exist in statically indeterminate
structures, originated from assembling / fabrication errors
of structural elements
A
1l 2l
3l
A
FN1 FN3 FN2
A
1 23
l
• Force equilibrium:
• Displacement compatibility:
N1 N2
N3 N1 N2( )cos
F F
F F F
cos
1
3
ll
• Bar 3 is δ shorter than required
Assembly Stress
21
• Given: AB rigid; E1A1 = E2A2 = EA; Bar 1 is δ shorter than required. Find: internal forces in Bar 1 and 2 after assembly.
• Solution:
Bar 1 lengthened & Bar 2
shortened:
2. Displacement compatibility:
11 2
2
22
l al l
l a
BA
21
1l
a a
2l
BA
F1 F2FA
1 20 2AM aF aF
1. Basic determinate structure and force equilibrium of AB:
Sample Problem
22
3. Displacement-force relationship:
1 21 2;
Fl F ll l
EA EA
4. Complementary equation:
1 21 22( ) 2
Fl F ll l
EA EA
1 2 1
1 2
2
42
5
2 2
5
EAaF aF F
lFl F l
EAFEA EA
l
5. Joint of force equilibrium and complementary equations
23
• A temperature change results in a change in length or
thermal strain. There is no stress associated with the
thermal strain unless the elongation is restrained by
the supports.
thermal expansion coefficient
T P
PLT L
AE
• Treat the additional support as redundant and apply
the principle of superposition.
0
0
AE
PLLT
PT
P AE T
PE T
A
• The thermal deformation and the deformation from
the redundant support must be compatible.
Thermo Stress
• Example (low-carbon steel)
6 0200 GPa 12.5 10 2.5 MPaC T T
24
Coefficient of Thermal Expansion
25
• Extension / contraction knots:
• Extension / contraction slot:
Addressing Thermo Stresses
26
Statically Indeterminate Shafts
• Determine (a) the reactive torques at the ends, (b) the maximum shearing stresses in each segment of the bar, and (c) the angle of rotation at the cross section where the load T0 is applied.
• Solution:
0
/
0 0
max max
0
,
2 2,
B AB BB A
pB pA
A pB B pA
B A
B pA A pB B pA A pB
B B A A B B A ABC AC C
pB pA pB pA
T T LT L
GI GI
L I L IT T T T
L I L I L I L I
T d T d T L T L
I I GI GI
27
Statically Indeterminate Shafts
• For the special case of dA = dB:
0 0 0 0
0 0
max max
0
,
2 2,
2 2
A pB B pAA BB A
B pA A pB B pA A pB
B AA BBC AC
p p p p
A BB B A AC
p p p
L I L IL LT T T T T T
L I L I L L I L I L
T d T dL dT L dT
I LI I LI
L L TT L T L
GI GI LGI
28
Sample Problem
• What is the deformation of the
rod and tube when a torque T is
exerted on a rigid end plate as
shown?
1 2
1 21 2
1 1 2 2
1 1 2 2
1 2
1 1 2 2 1 1 2 2
1 2
1 1 2 2
,
p p
p p
p p p p
p p
T T T
T L T L
G I G I
G I T G I TT T
G I G I G I G I
TL
G I G I
• Solution:
29
Sample Problem
• Determine the maximum shearing stress in each of
the two shafts with a diameter of 1.5 in.
• Solution:
4max
max max
4 12 600
2 12 0
2 2 4 4
120 lb ft, 240 lb ft, 1440 lb
2 240 12 1.5 24.35 ksi
1.5 32
2 12.17 ksi
2
A
B
B AF E
p p
A B
B
BD
p
A
AC BD
p
T F
T F
T L T L
GI GI
T T F
T d
I
T d
I
30
• Make the beam determinate by replacing the restraint at B with FB
l
q
A B
qA B
FB
qBA
MA• Make the beam determinate by replacing the rotational restraint at A with MA
• A propped cantilever beam
Statically Indeterminate Beams
31
q
AB
l
• Known: q and l. Find: reaction forces at the supports.
Sample Problem
32
• Method 1:
A B
q
FB
3 4
0
03 8
3
8
B
B
B
w
F l ql
EI EIql
F
• Deformation compatibility requires:
• Take the deflection restraint at B as the redundant restraint
33
A B
q
l
• Take the rotational restraint at A as the redundant restraint
q
BA
MA
3
2
0
03 24
8
A
A
A
M l ql
EI EI
qlM
• Deformation compatibility requires:
A B
q
l
• Method 2:
34
PA
B
C D
aa
• Cantilever beam AB is enhanced by another cantilever
beam CD with the same EI. Find (1) contact force at D;
(2) ratio of deflections at B with and without enhancement;
(3) ratio of the maximum bending moments in AB with and
without enhancement.
Sample Problem
35
• Contact force at DP
A
B
CFD
FD
• Deflection compatibility
at D requires D Dw AB w CD
3
3
DD
F aw CD
EI
CD
FD
• Solution:
36
2 32 35
3 3 26 6 3 6
D DD
F a F aPa Paw AB a a a a
EI EI EI EI
P
AB
D1
FD
AB
D
FD
P
AB
D
3 335 5
3 6 3 4
D DD D D
F a F aPaw AB w CD F P
EI EI EI
37
• Ratio of deflections at B with and without the enhancement
3 3
0
2 33 3 3
1
3 3
1
0
2 8
3 3
58 8 393 2
6 3 6 3 24
3939 8
24 3 64
B
D DB
B
B
P a Paw
EI EI
F a F aPa Pa Paw a a
EI EI EI EI EI
w Pa Pa
EI EIw
38
• Ratio of the maximum bending moments in AB with and
without the enhancement.
0 0max 0 2AM x Px M M Pa
• Without:
• With:
1
1max 1
2D
D
Px x aM x
Px F x a a x a
M M Pa
PA B
x
P
ABD
FD
x
39
Exercise
• Find the reaction forces for the following continuous beam.
MMq qa
a2a
AB D
C
a
q qa
a2a
AB D
C
a
• Solution
40
Moment-Area Theorems with Indeterminate Beams
• Reactions at supports of statically indeterminate
beams are found by designating a redundant
constraint and treating it as an unknown load which
satisfies a displacement compatibility requirement.
• The (M/EI) diagram is drawn by parts. The
resulting tangential deviations are superposed and
related by the compatibility requirement.
• With reactions determined, the slope and deflection
are found from the moment-area method.
41
• For the coplanar structure shown below, the flexural
rigidity of AB is EI, tension rigidity of CD is EA; P, L and
a are given. Find FCD.
P
A BC
D
a
2L
2L
Combined Indeterminate Structures
42
• Solution
3
3
3
48
48
C CD
C CC CD
C
w l
P F L F aw l
EI EA
ALF P
Ia AL
D
a
C
P
A BC
2L
2L
CF
• Deformation compatibility requires:
43
Combined Indeterminate Structures
More Examples
44
More Examples
More Examples
46
More Examples
47
Consider
only the
effects of
bending.
• Statically Determinate Problems(静定问题)
• Statically Indeterminate Problems(超静定问题)
• Degrees of Indeterminacy(超静定次数)
• Advantages & Disadvantages of Indeterminate Structures
(超静定结构的优缺点)
• Redundancy & Basic Determinate System(冗余约束与基本静定系)
• General Ideas for Analyzing Indeterminate Structures
(超静定问题一般解法)
Contents
48
• Force Method Solution Procedure(力法)
• Statically Indeterminate Bars(超静定拉压杆)
• Assembly Stress(装配应力)
• Thermo Stress(热应力)
• Statically Indeterminate Shafts(超静定扭转轴)
• Statically Indeterminate Beams(超静定弯曲梁)
• Moment-Area Theorems with Indeterminate Beams(图乘法求解超静定梁)
• Combined Indeterminate Structures (组合超静定结构)
Contents
49