Structures - Class Exercise 1a.
1.101 Sophomore Design - Fall 2006
Given the differential equation:
2
Qo
w
(
x
)
=
------ ⋅ ( L – x) on 0 ≤ x ≤ L where capitalized letters
2
EI
dx
d
dw
dx
are constants. The boundary conditions are
Find
w(x) = ?
dw
dx
= ?
= 0
and
w
x=0
= 0
x=0
and
w
x=L
= ?
x=L
Qo
2
3
From direct integration of the differential equation w ( x ) = ------ ⋅ ⎛ -----L – -----⎞ + c 1 ⋅ x + c 2
EI ⎝ 2
6⎠
x
x
From the boundary conditions, the two constants of integration must be zero, so
3
Qo ⎛ x2
x ⎞
w ( x ) = ------ ⋅ -----L – ----EI ⎝ 2
6⎠
and
2
Qo ⎛
d
x ⎞
w ( x ) = ----- ⋅ xL – ----dx
EI ⎝
2⎠
Then, at x= L, the deflection, w(L) = ∆, and the slope, dw
dx
3
Qo ⋅ L
∆ = w( L ) = ----------------3EI
and
= φ ( L ) are
x=L
d
φ ( L ) = (w)
dx
2
x=L
Qo ⋅ L
= ---------------2EI
What does all this have to do with the picture shown?
Qo is the end load. w(x) is the deflection. The differential
equation we start with is the “moment-curvature” rela
tion. For small deflections and rotations,
d
dx
2
2
w(x)
w ( x ) is the
curvature (positive concave upwards). The right hand
side is the bending moment which, by convention of
1.050, is taken as positive when the curvature is concave
downwards. M y = –Q o ⋅ ( L – x)
October 23, 2006
Qo
Deflection
∆
L
My
Bending Moment
x
M y = –Q o ⋅ ( L – x)
1
Structures - Class Exercise 1b.
1.101 Sophomore Design - Fall 2006
2
–M o
w
(
x
)
=
---------2
EI
dx
d
Given the differential equation:
letters are constants. The boundary conditions are
Find
w(x) = ?
dw
dx
= ?
on 0 ≤ x ≤ L where all capitalized
dw
dx
= 0
w
x=0
= 0
x=0
and
w
x=L
= ?
x=L
As in the previous exercise, we find, after applying the boundary conditions
2
–M o x
w ( x ) = ---------------2EI
–M o x
d
w ( x ) = -----------dx
EI
and
Then, at x= L, the deflection, w(L) = ∆, and the slope, dw
dx
= φ ( L ) are
x=L
2
–M o L
∆ = w ( L ) = ---------------2EI
and
φ(L) =
d
(w)
dx
x=L
–M o L
= ------------EI
What does all this have to do with the picture shown?
Deflection
Here, Mo is the applied moment at the end of the beam.
Within the beam, 0<x<L, the bending moment is con
stant and equal to the applied moment. It is positive
according to our convention.
L
w(x) ∆
My
My = Mo
Bending Moment
Mo
x
We now superimpose these two loading cases to solve a more general problem and one that is rel
evant to your design task.
October 23, 2006
2
For the cantilever under end load and end moment, the deflected shape might look as shown .
Qo
w(x)
∆
φ<0
Mo
L
For this combined loading, we have:
2
3
Qo ⎛ x2
x ⎞ –M o x
w ( x ) = ------ ⋅ -----L – ----- + --------------EI ⎝ 2
6⎠
2EI
2
Qo ⎛
d
x ⎞ –M o x
w ( x ) = ------ ⋅ xL – ----- + -----------dx
EI ⎝
2⎠
EI
At the end of the beam, we have:
3
2
Qo ⋅ L Mo L
∆ = ----------------- –------------3EI
2EI
2
Qo ⋅ L Mo L
φ ( L ) = ---------------- – ---------2EI
EI
and
We want the solution for the special case when the slope at x = L is zero.
Qo
w(x)
Mo
φ=0
∆
L
Qo ⋅ L
= ------------Setting φ(L) = 0 gives the applied end moment in terms of the end force M o
2
φ(L) = 0
3
Qo ⋅ L
So, in this case, the tip deflection is ∆ = ----------------- or, expressing Qo in terms of the displacement,
12EI
Qo = K ⋅ ∆
where the stiffness, K, is
October 23, 2006
12 ⋅ EI
K = --------------3
L
3
Engineering Beam Theory
Engineering beam theory shows that the most
significant stress is the normal stress component on an “x face”; σx in the example at the
right. It is related to the applied loads by
z
σ x , εx
L
P z,
w(x)
a
where z is the distance from the “neutral
axis” which, for a doubly symmetric beam, is
at the center of the cross-section and I is the
moment of inertia of the cross section.
∫z
2
dA
P
a
x
z
My.
A
My
x
My ⋅ z
σ x = -------------I
I =
Qz
For a rectangular cross-section of width b
and height h, this is I = bh3/12
My.
R
The applied loads come in through the bend
ing moment My. The convention for positive
shear and bending moment is shown in the
figure.
The extensional strain, from the stress/strain
relations is just ε x = σ x ⁄ E where E is the Elastic Modulus. In terms of the geometry of
deformation, the extensional strain is given by ε x = z ⁄ R where R is the radius of curvature of
the neutral axis. (1/R) is the curvature. The “bending stiffness” is defined as the product E I as it
appears in the “moment-curvature” relationship M y = ( EI ) ⋅ ⎛ -- ⎞
⎝ R⎠
1
where the curvature, for
small deflections, is related to the vertical displacement of the neutral axis by, (1/R) = - d2w/dx2
or .
– M y ⁄ ( EI ) =
d
2
2
w(x)
dx
An integration of the differential equation obtained from the moment curvature relation gives, for
the case where the beam is loaded as shown, the mid-span deflection
w
midspan
Pa
2
2
= – ⎛⎝ ------------⎞⎠ ⋅ ( 3L – 4a )
24EI
October 23, 2006
4