Statically Indeterminant Beams
If we have
h
too
t many supports
t or supportt conditions
diti
a beam
b
will
ill b
be
statically indeterminant and we will not be able to solve for the
external reactions or the internal forces. Example:
A
w lb/length
B
L
w
MB
A
B
(two equations of equilibrium, three unknown reactions)
Just as in the case of springs or trusses, we can solve statically
indeterminant problems if we relate the loads to the deformation
In a beam the vertical deflection of the beam, v(x), is related to
the internal bending moment, M(x), in the beam, through the
relation
d 2v
EI 2 = M ( x )
dx
y
v(x)
x
E … Young's modulus of the beam (a material property)
I … an area moment of the beam cross sectional Area, A ( a geometry
property)
p
p y)
2
I = ∫ y dA
A
We can use this relationship and appropriate boundary conditions on
th beam
the
b
slope
l
or vertical
ti l deflection
d fl ti to
t find
fi d allll th
the reactions
ti
and
d make
k
the problem solvable.
pin or roller support
v =0 (no deflection)
d /d = slope
dv/dx
l
iis nott
constrained
fixed (clamped) support
v=0 (no deflection)
dv/dx =0 (no slope)
Consider our previous example
y
w lb/length
B
A
x
L
wx
x/2
M(x)
A
x
M(x) = Ax –wx2/2
V(x)
d 2v
wx 2
EI 2 = Ax −
dx
2
dv Ax 2 wx3
EI
=
−
+ C1
2
6
dx
Ax 3 wx 4
EIv =
−
+ C1 x + C2
6
24
v ( 0 ) = 0 ⇒ C2 = 0
AL3 wL4
v ( L) = 0 ⇒
−
+ C1 L = 0
6
24
dv
AL2 wL3
−
+ C1 = 0
( L) = 0 ⇒
dx
2
6
Solving
g for A,, C1
3
A = wL
8
1
C1 = − wL3
48
knowing A, we can
solve for B and MB
w
MB= -wL2/8
B=5wL/8
A = 3wL/8
We can also now
look at the beam
deflection
3wLx 3 wx 4 wL3
EIv ( x ) =
−
−
x
48
24
48
3
4
EI
3 ⎡x⎤
1 ⎡x⎤
1 ⎡x⎤
=
−
−
v
x
(
)
48 ⎢⎣ L ⎥⎦ 24 ⎢⎣ L ⎥⎦ 48 ⎢⎣ L ⎥⎦
wL4
deflected shape (exaggerated)
If we use singularity functions to write the internal bending moment we
can solve statically indeterminant problems with various loading
conditions relatively easy
500 lb
y
x
A
6'
B
3'
C
3'
from force and moment equilibrium
500
A
C=250 +A
B = 250-2A
M ( x ) = A x − 0 + B x − 6 − 500 x − 9 + C x − 12
1
1
1
1
d 2v
M ( x ) = EI 2 = A
dx
dv A
2
=
EI
x−0 +
dx 2
A
B
3
EIv =
x−0 +
6
6
x − 0 + B x − 6 − 500 x − 9 + C x − 12
1
1
1
B
C
2
2
2
x − 6 − 250 x − 9 +
x − 12 + C1
2
2
250
C
3
3
3
x−6 −
x−9 +
x − 12 + C1 x + C2
3
6
The
e bou
boundary
da y co
conditions
d o sa
are
e
v(0) =0
C2 =0
v(6) =0
6A +6C1 = 0
v(12) =0
1
note: could drop this term throughout
since it does not contribute to deflection
or slope for 0<x<12. It is only really
needed in the shear force expression
288 A + 36 B − 2250 + 12C1 = 0
using the relationship between B and A we can
solve for A and C1 . Solving for B and C then
gives
A = -33.1
33 1 lb
B =316.2 lb
C= 216.9 lb