11
Beam Deflections:
4th Order Method
and Additional Topics
11–1
Lecture 11: BEAM DEFLECTIONS: 4TH ORDER METHOD AND ADDITIONAL TOPICS
TABLE OF CONTENTS
Page
§11.1.
§11.2.
§11.3.
Fourth Order Method Description
§11.1.1. Example 1: Cantilever under Triangular Distributed Load . . .
Superposition
§11.2.1. Example 2: Cantilever Under Two Load Cases . . . . . .
§11.2.2. Example 3: A Statically Indeterminate Beam . . . . . . .
Continuity Conditions
§11.3.1. Example 4: Simply Supported Beam Under Midspan Point Load
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11–3
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11–4
11–4
11–5
11–6
11–6
§11.1
FOURTH ORDER METHOD DESCRIPTION
§11.1. Fourth Order Method Description
The fourth-order method to find beam deflections gets its name from the order of the ODE to be
integrated: E Izz v I V (x) = p(x) is a fourth order ODE. The procedure can be broken down into the
following steps.
1.
Express the applied load p(x) as function of x, using positive-upward convention. This step
may involve changing load signs as necessary, as in the example below.
2–3. Integrate p(x) twice to get Vy (x) and Mz (x)
4.
Pause. Determine integration constants from static BCs, and replace in Mz (x). (If the constants
are too complicated when expressed in terms of the data, they might be kept in symbolic form
until later.)
5–8. From here on, same as the second order method.
An example of this technique follows.
§11.1.1. Example 1: Cantilever under Triangular Distributed Load
This example has been worked out in the previous lecture using the second order method. It is
defined in Figure 11.1, which reproduces the figure of the previous Chapter for convenience.
Constant EIzz
y
w(x) = wB x /L
wB
B
x
A
L
Figure 11.1. Beam problem for Example 1. The applied load w(x) = w B x/L is considered positive if it goes
downward, that is, if w B > 0. This is converted to a negative load p B (x) = −w B x/L to insert in the ODEs.
From inspection the applied load is
wB x
.
(11.1)
L
Notice the minus sign to pass from the user’s convention: w(x) > 0 if directed downward, to the
generic load convention: p(x) > 0 if directed upward. Integrating p(x) twice yields
p(x) = −
wB x 2
+ C1 ,
2L
wB x 3
Mz (x) = − Vy (x) d x = −
− C1 x + C2 .
6L
Vy (x) = −
p(x) d x =
(11.2)
Apply now the static BCs at the free end A: Vy A = Vy (0) = C1 = 0 and Mz A = Mz (0) = C2 = 0.
Hence
wB x 3
(11.3)
Mz (x) = − 12 w(x) x ( 13 x) = −
6L
11–3
Lecture 11: BEAM DEFLECTIONS: 4TH ORDER METHOD AND ADDITIONAL TOPICS
Constant EIzz
w(x) = wB x /L
y
A
wB
B
x
L
P
(a) Original problem
y
=
w(x) = wB x /L
y
B
x
A
+
wB
B
x
A
L
L
P
(b) Decomposition into two load cases and superposition
Figure 11.2. Beam problem for Example 2.
From here on the steps are the same as in the second order method worked out in Lecture 10. The
deflection curve is
wB
(x 5 − 5L 4 x + 4L 5 )
v(x) = −
(11.4)
120E Izz L
The maximum deflection occurs at the cantilever tip A, and is given by
v A = v(0) = −
wB L 4
⇓
30E Izz
(11.5)
The negative sign indicates that the beam deflects downward if w B > 0.
§11.2. Superposition
All equations of the beam theory we are using are linear. This makes possible to treat complicated
load cases by superposition of the solutions of simpler ones. Simple beam configurations and load
cases may be compiled in textbooks and handbooks; for example Appendix D of Beer-JohnstonDeWolf. The following example illustrates the procedure.
§11.2.1. Example 2: Cantilever Under Two Load Cases
Consider the problem shown in Figure 11.2(a). The cantilever beam is subject to a tip point force
as well as a triangular distributed load. This combination can be decomposed into the two load
cases shown in Figure 11.2(b). Both of these have been separately solved previously as Examples
1 and 2 of Lecture 10 (the latter also as the example in the previous section). The deflection curves
for these cases will be distinguished as v P (x) and v w (x), respectively.
We had obtained
v P (x) = −
P
(L − x)2 (2L + x),
6E Izz
v w (x) = −
11–4
wB
(x 5 − 5L 4 x + 4L 5 )
120E Izz L
(11.6)
§11.2
Constant EIzz
y
A
w(x) = wB x /L
wB
B
x
A
L
(a) Original (statically indeterminate) problem
B
+
w(x) = wB x /L
A
wB
B
x
L
RA
RB
(b) Support reactions
y
x
MB
B
RA
y
= A
SUPERPOSITION
L
(c) Decomposition into two load cases and superposition
Figure 11.3. Beam problem for Example 3.
The deflection under the combined loading is obtained by adding the foregoing solutions:
v(x) = v P (x) + v w (x) = −
P
wB
(x 5 − 5L 4 x + 4L 5 ).
(L − x)2 (2L + x) −
6E Izz
120E Izz L
(11.7)
The tip deflection is
v A = v(0) = −
P L3
wB L 4
L3
−
=−
(10P + w B L)
3E Izz
30E Izz
30E Izz
⇓
(11.8)
Superposition can be also used for any other quantity of interest, for example transverse shear
forces, bending moments and deflection curve slopes. An application to statically indeterminate
beam analysis is given next.
§11.2.2. Example 3: A Statically Indeterminate Beam
The problem is defined in Figure 11.3(a). The beam is simply supported at A and clamped at B. If
the supports are removed 3 reactions are activated: R A , R B and M B , as pictured in Figure 11.3(b).
Many point = 0
But thereare only two nontrivial static equilibrium equations:
Fy = 0 and
because
Fx = 0 is trivially satisfied. Consequently the beam is statically indeterminate because
the reactions cannot be determined by statics alone. One additional kinematic equation is required
to complete the analysis.
We select reaction R A as redundant force to be carried along as a fictitious applied load. Removing
the support at A and including R A makes the beam statically determinate. See Figure 11.3(b). This
beam may be viewed as being loaded by a combination of two load cases: (1) the actual triangular
load w(x), and (2) a point load R A at A. But this is exactly the problem solved in Example 2, if we
replace P by −R A . The deflection curve of this beam is
v(x) =
wB
RA
(x 5 − 5L 4 x + 4L 5 ).
(L − x)2 (2L + x) −
6E Izz
120E Izz L
11–5
(11.9)
Lecture 11: BEAM DEFLECTIONS: 4TH ORDER METHOD AND ADDITIONAL TOPICS
Now the tip deflection must be zero because there is a simple support at A. Setting v A = v(0) = 0
provides the value of R A :
RA L3
wB L 4
−
=0
v A = v(0) =
3E Izz
30E Izz
⇒
RA =
wB L
10
(11.10)
This reaction value can be substituted to complete the solution. For example, the bending moment
is
wB x 3
wB L x
wB x 3
wB x
Mz (x) = R A x −
(11.11)
=
−
=
(3L 2 − 5x 2 )
6L
10
6L
30L
√
The moment is zero at A (x = 0), becomes positive for 0 < x < L 3/5 ≈ .7746 L, crosses zero
at x = 0.7746 L and reaches Mz B = −w B L 2 /15 at the fixed end. The deflection is
v(x) =
wB
wB L
(x 5 − 5L 4 x + 4L 5 ),
(L − x)2 (2L + x) −
60E Izz
120E Izz L
(11.12)
which may be simplified to
v(x) = −
wB
x (L 2 − x 2 )2
120 E Izz L
(11.13)
§11.3. Continuity Conditions
If the applied load is discontinuous, i.e., not a smooth function of x, it is necessary to divide the
beam into segments separated by the discontinuity points. The ODEs are integrated over each
segment. These solutions are “patched” by continuity conditions expressing that the slope v (x)
and the deflection v(x) are continuous between segments. This matching results in extra relations
between integration constants, which permits elimination of all integration constants except those
that can be determined by the standard BCs. The procedure is illustrated with the next example.
§11.3.1. Example 4: Simply Supported Beam Under Midspan Point Load
The problem is defined in Figure 11.4(a). The calculation of the deflection curve will be done by the
second order method. Divide the beam into two segments: AC, which extends over 0 ≤ x ≤ 12 L,
and CB, which extends over 12 L ≤ x ≤ L. For brevity, these are identified as segments 1 and 2,
respectively, in the equations below.
The expression of the bending moment over each segment is easily obtained from statics. From
symmetry, the support reactions are obviously R A = R B = 12 P as shown in Figure 11.4(b). By
inspection one obtains that the bending moment Mz (x), diagrammed in Figure 11.4(c), is

 Mz1 (x) = P x
over segment 1 (AC),
2
(11.14)
Mz (x) =
 M (x) = P(L − x) over segment 2 (CB).
z2
2
11–6
§11.3
y
A
Constant EIzz
P
x
P
A
B
C
L/2
CONTINUITY CONDITIONS
C segment 2
segment 1
RA = P
2
L/2
L
(a) Problem definition
B
RB = P
2
(b) Support reactions and division
into two segments
P
A
RA = P
2
segment 1
C segment 2
B
RB = P
2
Mz2 (x)= P(L-x)
2
Mz1 (x)= Px
2
+
+
(c) Bending moment diagram
Figure 11.4. Beam problem for Example 4.
Integrate Mz /E Izz over each segment:

2
 E Izz v (x) = P x + C1
1
4
E Izz v (x) =
 E I v (x) = P x (2L − x) + Ĉ
zz 2
1
4
over segment 1 (AC),
(11.15)
over segment 2 (CB).
It is convenient to stop here and get rid of Ĉ1 to avoid proliferation of integration constants. To
do that, note that the midspan slope vC must be the same from both expressions: vC = v1 ( 12 L) =
v2 ( 12 L). Else the beam would have a “kink” at C. This is called a continuity condition. Equating
P L 2 /16 + C1 = (3/16)P L 2 + Ĉ1 yields Ĉ1 = C1 − P L 2 /8, which is replaced in the second
expression above:

2
 E Izz v (x) = P x + C1
over segment 1 (AC),
1
4
E Izz v (x) =
(11.16)

− x) − P L 2 + C over segment 2 (CB).
E Izz v2 (x) = P x (2L
1
4
8
Now Ĉ1 is gone. Integrate again both segments:

3
 E Izz v1 (x) = P x + C1 x + C2
12
E Izz v(x) =
2

P
− x) − P L 2 x + C x + Ĉ
E Izz v2 (x) = x (3L
1
2
12
8
over segment 1 (AC),
over segment 2 (CB).
(11.17)
To get rid of Ĉ2 we say that the midspan deflection vC must be the same from both expressions:
vC = v1 ( 12 L) = v2 ( 12 L). This continuity condition gives Ĉ2 = C2 + P L 3 /48, which replaced
11–7
Lecture 11: BEAM DEFLECTIONS: 4TH ORDER METHOD AND ADDITIONAL TOPICS
yields

3
 E Izz v1 (x) = P x + C1 x + C2
12
E Izz v(x) =
2

P
P L2 x + P L3 + C x + C
−
E Izz v2 (x) = x (3L−x)
1
2
12
8
48
over segment 1 (AC),
over segment 2 (CB).
(11.18)
We have now only two integration constants. To determine C1 and C2 use the kinematic BCs at A
and B. v A = v1 (0) = C2 = 0 and v B = v2 (L) = 0 ⇒ C1 = −P L 2 /16. Substitution gives, after
some simplifications,

 E Izz v1 (x) = P x (4x 2 − 3L 2 )
48
E Izz v(x) =
 E Izz v2 (x) = − P (4x 3 − 12L x 2 + 9L 2 x − L 3 )
48
over segment 1 (AC),
over segment 2 (CB).
(11.19)
The midspan deflection, obtainable from either segment, is
vC = v1 ( 12 L) = v2 ( 12 L) = −
P L3
48E Izz
(11.20)
As can be seen the procedure is elaborate and error prone, even for this very simple problem. It can
be streamlined by using Discontinuity Functions (DFs), which are covered in Lecture 12.
11–8