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Material Type: Notes; Class: HONORS ENGR STATICS; Subject: ENGINEERING MECHANICS; University: Iowa State University; Term: Unknown 1989;
Typology: Study notes
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Statically Indeterminant Beams
If
h
t
t
t
diti
b
ill b
If we have too many supports or support conditions a beam will bestatically indeterminant and we will not be able to solve for theexternal reactions or the internal forces. Example:
w lb/length
w
B
(two equations of equilibrium, three unknown reactions)
Just as in the case of springs or trusses, we can solve staticallyindeterminant problems if we relate the loads to the deformationindeterminant problems if we relate the loads to the deformationIn a beam the vertical deflection of the beam, v(x), is related tothe internal bending moment, M(x), in the beam, through therelation
2
v(x)
2
y
x
… Young's modulus of the beam (a material property)
I
… an area moment of the beam cross sectional Area, A ( a geometry
property)
2
p
p
y)
2
A
Consider our previous example
w lb/length
y
x
wx
x/
V(x)
M(x)
x
M(x) = Ax –wx
2
2
2
2
2
3
1
3
4
1
2
(
)
(
)
2
3
4
1
2
3
0
0
0
0
0
6
24
v
C
AL
wL
v L
C L
=
⇒
=
=
⇒
−
=
(
)
2
3
1
0
0
2
6
dv
AL
wL
L
C
dx
=
⇒
−
=
Solving for A, C
1
g
1
(^38)
A
wL
=
3
1
1 48
C
wL
= −
If we use singularity functions to write the internal bending moment wecan solve statically indeterminant problems with various loadingcan solve statically indeterminant problems with various loadingconditions relatively easy
500 lb
y
x
from force and moment equilibrium
(
)
1
1
1
1
0
6
500
9
12
M
x
A x
B x
x
C x
=
−
−
−
−
−
(
)
2
1
1
1
1
2
0
6
500
9
12
d v
M
x
EI
A x
B x
x
C x
dx
=
=
−
−
−
−
−
2
2
2
2
1
3
3
3
3
1
2
0
6
250
9
12
2
2
2
250
0
6
9
12
dv
A
B
C
EI
x
x
x
x
C
dx
A
B
C
EIv
x
x
x
x
C x
C
=
−
−
−
−
−
=
−
−
−
−
−
1
2
0
6
9
12
6
6
3
6
EIv
x
x
x
x
C x
C
The boundary conditions are
note: could drop this term throughout
e bou da y co d
o s a e
v(0) =
2
note: could drop this term throughoutsince it does not contribute to deflectionor slope for 0<x<12. It is only reallyneeded in the shear force expression
v(6) =
1
288
36
2250
12
0
A
B
C
v(12) =
1
288
36
2250
12
0
A
B
C
−
=
33 1 lb
using the relationship between B and A we can
A = -33.1 lbB =316.2 lbC= 216.9 lb
solve for A and C
1
. Solving for B and C then
gives
