Lecture 6
Moderately Large Deflection Theory of Beams
Problem 6-1:
Part A: The department of Highways and Public Works of the state of California is in the
process of improving the design of bridge overpasses to meet earthquake safety criteria. As a
highly paid consultant to the project, you were asked to evaluate its soundness. You rush back to
your lecture notes, and you model the overpass as a simply supported beam of span L with an
overhang =0.01L. Assume that the distributed load is a sinusoidal function.
EI
EA
N
L
a) Calculate the maximum allowable midspan deflection (wo)critical under which the beam
will slide off its support.
Part B: Assume that the above design with an external axial force N=0 and =0.01L has a
safety factor of one. The design of earthquake resistant structures requires a safety factor of five,
meaning that (wo)critical must be increased by a factor of five without the bridge collapsing. Two
possible design modifications were proposed. In the first one, the overhang is simply increased to
In the second design, a tensile force N is applied to the bridge to increase its transverse
new.
stiffness and thus reduce the central deflection and the resulting motion of the support.
b) For the first proposed modification, what length new of the overhang will meet the
requirement of a safety factor of five? Give your result in terms of the original and
other parameters if needed.
c) For the second design, what is the magnitude of the dimensionless tensile force N/EA that
will give a safety factor equal to five?
d) Which design is better? Can you think of a third alternative design solution?
1
Problem 6-1 Solution:
Recall:
NL
EA
(a) Calculate the max deflection
L
0
wo
artical
2
1 dw
2 dx
dx
at L / 2 under which the beam will slide off its
support
Assume N
0
L
0
2
1 dw
2 dx
dx
(1)
Since the applied load is a sin function, we can assume the deflected shape will also be a
sine function
w' x
2
w' x
x
L
wo sin
w x
wo
L
x
L
cos
2
wo
2
cos2
L
x
L
Substitute the above equation into equation (1), we have
L
0
1 2
wo
2
L
1 2
wo
2
L
2
1 2
wo
2
L
2
1 wo
L 2
2
cos 2
x
dx
L
sin x L
4 L
x
2
L
0
L
0
2
2
1 wo
L 2
2
2
Thus, the maximum allowable midspan deflection is
wo 2
2
wo
wo
new
new
1 wo
L 2
L2
100
L
5
L
5
L
5
to meet safety factor of 5
increase
5wo
Recall:
L
0.01L
wo
(b) Case 1
2
2
wo
We’re given
4L
1 5wo
L
2
2
L
new
2
1 5wo
L
2
2
25
1 wo
L 2
2
2
new
25
(c) Case 2: apply a tensile force N, so now N
NL
EA
NL
EA
0
L
0
2
1 dw
2 dx
1 wo
L 2
dx
2
3
We want to calculate N that will give us the equivalent effect of applying a safety factor
of 5, in which case
1 wo
L 2
NL
EA
In the case of
2
25
25
24
24 L
L 100
0.24
L /100
N
EA
(d) Which design is better?
It is difficult to say which design is better. Each design has its advantage and
disadvantages. For case 1, we will have a long overhang which may not be aesthetically
pleasing. For case 2, it may be difficult to apply constantly a tensile force.
Other options include a stiffer simply supported beam, or add cables to suspend the
bridge.
4
Problem 6-2:
A long span aerial tramway steel cable of length L=1km is loaded by a hurricane wind with
intensity q(x) sinusoidally distributed between the end stations. The cable deflects by wo=5m.
2.1 105 MPa
E
y
Cross-section
of cable
D
300MPa
D 60mm
q( x) q0 sin
x
L
w0
q0
q(x)
L
a) Calculate the resulting load intensity qo
b) Calculate the tension in the cable N.
c) Calculate the tensile stress.
d) Compare (c) with the yield stress, and determine the safety factor.
Problem 6-2 Solution:
Using the equation of equilibrium
EIw
Nw'' q
However, a cable has no bending stiffness, so our equation becomes:
Nw '' q
w ''
q
N
1
qo sin
N
x
L
Integrate twice
w'
w
qo L
cos
N
qo L
N
2
sin
x
L
x
L
C1
C1 x C2
5
Plug in the boundary conditions to solve for the constants
w 0
0
w 0
C2
w L
0
w L
0 C1 L C2
2
qo L
N
w x
a)
0
0
0
C1
x
L
sin
Calculate the load intensity
The cable deflects by wo
5m at the middle point x
L
w
2
qo L
N
wo
2
L/2
qo L
N
L
sin
L2
2
2
Nwo
qo
b)
L
Calculate the tension on the cable(N)
N
Using
EA
1
L
L
0
1 dw
2 dx
2
dx and w x
qo L
N
w' x
N3
EA
2L
L
0
2
x
L
cos 2
EAL qo
2 N
2
EAL qo
2 N
2
EA qo L
4
cos 2
2
x
2
x
L
sin
x
L
2
qo L
N
2
cos
L
qo L
N
2
w' x
N
qo L
N
x
dx
L
sin 2x L
4 L
L
0
L
0
2
2
6
2
qL
EA o
2
N
2
Given wo
5m , then qo
Nwo
N
c)
EA qo L
4
N
25EA
4
L
N
36626N
2
5N
L
2
EA
4
3
2
L
2
2
3
1
5N
L
25
2.1 105
4
L
10
25EA L
4
6
4
60 10
2
4
N
2
L
2
3 2
1 10 3
Calculate the tension stress on the cable
N
A
36626
60 10
3 2
12.95MPa
4
12.95MPa
d)
Compare with the yield stress and determine the safety factor
safety factor
yield stress
working stress
safety factor
23.17
300
12.95
7
Problem 6-3:
Plot the dimensionless deflections (wo/L) versus the dimensionless line load for both bending and
membrane (cable) solutions over a slender beam. At what dimensionless deflections will the
bending and membrane solutions be equal, assuming a length to thickness ratio equal to 10?
Problem 6-3 Solution:
Recall bending and membrane solutions:
Pure Bending
Po
w x
4
EI
at x
w
wo
wo
L
x
L
L
2
Po
w x
2
N
L
L
2
Po
EI
Po L
EI
w
4
Po L
h4
E
12
L
Po
wo
2
N
L
wo
h4
12
Po L
Eh 2
L
2
4
L
12L2
4 2
h
Po L
Eh 2
12L2
4 2
h
Eh
2
Po L
L
2
qo L
2
2
1 2
E qo Lh
2
where qo
3
Po
Eh
L
1
2
Po L
N
3
(Problem 6-2)
2
qo L
2
1
2
qL
EA o
2
so N
wo
L
2
Po L
N
where N
4
x
L
sin
L
2
at x
wo
where I
wo
sin
Membrane
2
1
1
2
3
3
Po
rearrange the above expression
wo
L
Po L
Eh 2
1
3
4
1
3
4
8
Let’s call
wo
L
y,
Po L
Eh 2
x
We want to plot
Bending y
Membrane y
12L2
x
4 2
h
4
1
3
4
x
1
3
Use a length to thickness ratio equal to 10:
L
10
h
Bending y 12.32 x
Membrane y
0.345 x
1
3
9
At what dimensionless deflections will the bending and membrane solutions be equal?
wo
L
bending
12.32x
x
wo
L
So at
Po L
Eh 2
deflections
wo
L
membrane
0.345 x
1
3
0.005
12.32 0.005 0.0577
0.005 , the bending and membrane solutions will be equal, where the dimensionless
wo
L
0.0577
10
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2.080J / 1.573J Structural Mechanics
Fall 2013
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