LECTURE
Third Edition
BEAMS: DEFORMATION BY
SUPERPOSITION
• A. J. Clark School of Engineering •Department of Civil and Environmental Engineering
19
Chapter
9.7 – 9.8
by
Dr. Ibrahim A. Assakkaf
SPRING 2003
ENES 220 – Mechanics of Materials
Department of Civil and Environmental Engineering
University of Maryland, College Park
LECTURE 19. BEAMS: DEFORMATION BY SUPERPOSITION (9.7 – 9.8)
Deflection by Superposition
„
Slide No. 1
ENES 220 ©Assakkaf
Method of Superposition
– When a beam is subjected to several loads
(see Fig. 18) at various positions along the
beam, the problem of determining the
slope and the deflection usually becomes
quite involved and tedious.
– This is true regardless of the method used.
– However, many complex loading
conditions are merely combinations of
1
Slide No. 2
LECTURE 19. BEAMS: DEFORMATION BY SUPERPOSITION (9.7 – 9.8)
ENES 220 ©Assakkaf
Deflection by Superposition
„
Method of Superposition
relatively simple loading conditions
P
y
w1
L
Figure 18
P
y w2
x
x
a
(a)
b
(b)
LECTURE 19. BEAMS: DEFORMATION BY SUPERPOSITION (9.7 – 9.8)
Deflection by Superposition
„
Slide No. 3
ENES 220 ©Assakkaf
Method of Superposition
– Assumptions:
• The beam behaves elastically for the combined
loading.
• The beam also behaves elastically for the each
of the individual loads.
• Small deflection theory.
2
LECTURE 19. BEAMS: DEFORMATION BY SUPERPOSITION (9.7 – 9.8)
Deflection by Superposition
„
Slide No. 4
ENES 220 ©Assakkaf
Method of Superposition
If it is assumed that the beam behaves
elastically for the combined loading, as well
as for the individual loads, the resulting final
deflection of the loaded beam is simply the
sum of the deflections caused by each of the
individual loads.
LECTURE 19. BEAMS: DEFORMATION BY SUPERPOSITION (9.7 – 9.8)
Deflection by Superposition
„
Slide No. 5
ENES 220 ©Assakkaf
Method of Superposition
– This sum may be an algebraic one (Figure
19) or it might be a vector sum as shown in
Figure 20, the type depending on whether
or not the individual deflection lie in the
same plane.
– The superposition method can illustrated
by various practical examples.
3
Slide No. 6
LECTURE 19. BEAMS: DEFORMATION BY SUPERPOSITION (9.7 – 9.8)
ENES 220 ©Assakkaf
Deflection by Superposition
„
Method of Superposition
y
P
Figure 19
w1
y
||
y
L
y
P
+
P
a
x
y
P
y w2
||
b
x
P
+w
x
u
x
x
y
wt
x
LECTURE 19. BEAMS: DEFORMATION BY SUPERPOSITION (9.7 – 9.8)
Method of Superposition
+
x
Slide No. 7
ENES 220 ©Assakkaf
Principle of Superposition:
• Deformations of beams subjected to
combinations of loadings may be
obtained as the linear combination of
the deformations from the individual
loadings
• Procedure is facilitated by tables of
solutions for common types of
loadings and supports.
4
Slide No. 8
LECTURE 19. BEAMS: DEFORMATION BY SUPERPOSITION (9.7 – 9.8)
Deflection by Superposition
„
ENES 220 ©Assakkaf
Method of Superposition
Figure 20
w
y
A
δy
P
A
δ = δ y2 + δ z2
δz
z
Slide No. 9
LECTURE 19. BEAMS: DEFORMATION BY SUPERPOSITION (9.7 – 9.8)
Deflection by Superposition
„
ENES 220 ©Assakkaf
Illustrative Example for the Use of
Superposition
– Consider the beam shown in Fig. 21, with
a flexural rigidity of EI = 100 MN⋅m.
y
150 kN
2m
Figure 21
20 kN/m
x
•D
L=8m
5
LECTURE 19. BEAMS: DEFORMATION BY SUPERPOSITION (9.7 – 9.8)
Deflection by Superposition
„
Slide No. 10
ENES 220 ©Assakkaf
Illustrative Example for the Use of
Superposition
– If we are interested on finding the slope
and the deflection, say of point D, then we
can use the superposition method to do
that as illustrated in the following slides.
– First we find the slope and deflection due
the effect of each load, i.e., w, P, etc.
LECTURE 19. BEAMS: DEFORMATION BY SUPERPOSITION (9.7 – 9.8)
Deflection by Superposition
„
Slide No. 11
ENES 220 ©Assakkaf
Illustrative Example for the Use of
Superposition
– The resulting final slope and deflection of
point D of the loaded beam is simply the
sum of the slopes and deflections caused
by each of the individual loads as shown in
Figure 22.
– We need to find both the slope and
deflection caused by the concentrated load
(120 kN) and distributed load (20 kN/m)
6
Slide No. 12
LECTURE 19. BEAMS: DEFORMATION BY SUPERPOSITION (9.7 – 9.8)
ENES 220 ©Assakkaf
Deflection by Superposition
„
y
Illustrative Example for the Use of
Superposition
150 kN
y
2m
20 kN/m
=
x
D
150 kN
y
2m
+
x
D
L=8m
20 kN/m
x
D
L=8m
L=8m
Figure 22. Original Loading is Broken into Two Individual Loads
Slide No. 13
LECTURE 19. BEAMS: DEFORMATION BY SUPERPOSITION (9.7 – 9.8)
ENES 220 ©Assakkaf
Deflection by Superposition
Illustrative Example for the Use of
150 kN Superposition150 kN
y
y
„
y
2m
2m
2m
20 kN/m
x
D
x
D
L=8m
δD
=
L=8m
δP
+
20 kN/m
x
D
L=8m
δw
δ D = δ D −due to P + δ D −due to w
Figure 23. Original Deflection is Broken into Two Individual Deflections
7
Slide No. 14
LECTURE 19. BEAMS: DEFORMATION BY SUPERPOSITION (9.7 – 9.8)
ENES 220 ©Assakkaf
Deflection by Superposition
„
Illustrative Example for the Use of
Superposition
– Slope and Deflection caused by P
• By either the direct integration or the singularity
functions method, it can be seen that the slope
and deflection (due to P) of point D of this
particular loaded beam are given, respectively,
as
2
3PL3
(θ D )P = − PL
and ( y D )P =
32 EI
256 EI
LECTURE 19. BEAMS: DEFORMATION BY SUPERPOSITION (9.7 – 9.8)
Deflection by Superposition
„
Slide No. 15
ENES 220 ©Assakkaf
Illustrative Example for the Use of
Superposition
– Slope and Deflection caused by P
• Therefore,
(θ D )P = −
PL2 150 × 103 (8) 2
=
= −0.003 rad
32 EI 32(100 × 106 )
(25a)
( yD )P = −
3PL3
3 150 × 103 (8)
=
= −0.009 m
256 EI 256(100 × 106 )
(25b)
(
)
3
8
Slide No. 16
LECTURE 19. BEAMS: DEFORMATION BY SUPERPOSITION (9.7 – 9.8)
Deflection by Superposition
„
ENES 220 ©Assakkaf
Illustrative Example for the Use of
Superposition
– Slope and Deflection caused by w
• By either the direct integration or the singularity
functions method, it can be seen that the slope
and deflection (due to w) of point D of this
particular loaded beam are given, respectively,
as
(26a)
(θ D )P = w (− 4 x 3 + 6 Lx 2 − L3 )
24 EI
(26b)
( yD )P = w (− x 4 + 2 Lx 3 − L3 x )
24 EI
Slide No. 17
LECTURE 19. BEAMS: DEFORMATION BY SUPERPOSITION (9.7 – 9.8)
Deflection by Superposition
„
ENES 220 ©Assakkaf
Illustrative Example for the Use of
Superposition
– Slope and Deflection caused by w
• With w = 20 kN/m, x = 2 m, and L = 8 m, thus
(θ D )P =
w
20 ×103
(− 356) = −0.00293 rad
− 4 x 3 + 6 Lx 2 − L3 =
24 EI
24 100 × 106
( yD )P =
w
20 × 10
(− 912) = −0.0076 m
− x 4 + 2 Lx 3 − L3 x =
24 EI
24 100 ×106
(
(
)
)
(
)
3
(
)
9
Slide No. 18
LECTURE 19. BEAMS: DEFORMATION BY SUPERPOSITION (9.7 – 9.8)
ENES 220 ©Assakkaf
Deflection by Superposition
Illustrative Example for the Use of
Superposition
„
– Combining the slopes and deflections
produced by the concentrated (P) and
distributed (w) loads, the results are
θ D = (θ D )P + (θ D )w = −0.003 − 0.00293 = −0.00593 rad
y D = ( y D )P + ( y D )w = −0.009 − 0.0076 = 0.0166 m = 16.6 mm
Slide No. 19
LECTURE 19. BEAMS: DEFORMATION BY SUPERPOSITION (9.7 – 9.8)
ENES 220 ©Assakkaf
Deflection by Superposition
„
y
Illustrative Example for the Use of
Superposition
150 kN
2m
y
20 kN/m
y
2m
x
D
x
D
L=8m
δD
θD
150 kN
=
(θ D )P
L=8m
(δ D )P
+
2m
20 kN/m
x
D
L=8m
(δ D )w
θ D = (θ D ) p + (θ D )w and y D = ( y D ) p + ( y D )w
(θ D )w
Figure 24. Total Slope and Deflection of Point D
10
LECTURE 19. BEAMS: DEFORMATION BY SUPERPOSITION (9.7 – 9.8)
Deflection by Superposition
„
Slide No. 20
ENES 220 ©Assakkaf
General Procedure of Superposition
– It is evident from the last results that the
slope or deflection of a beam is the sum of
the slopes or deflections produced by the
individual loads.
– Once the slopes or deflections produced
by a few typical individual loads have been
determined by one of the methods already
LECTURE 19. BEAMS: DEFORMATION BY SUPERPOSITION (9.7 – 9.8)
Deflection by Superposition
„
Slide No. 21
ENES 220 ©Assakkaf
General Procedure of Superposition
– Presented, the superposition method
provides a means of quickly solving a wide
range of more complicated problems by
various combinations of known results.
– As more data become available, yet a
wider range of problems can be solved by
the method of superposition.
11
LECTURE 19. BEAMS: DEFORMATION BY SUPERPOSITION (9.7 – 9.8)
Deflection by Superposition
„
Slide No. 22
ENES 220 ©Assakkaf
Slope and Deflection Tables
– To facilitate the task of practicing
engineers, most structural and mechanical
handbooks include tables giving the
deflections and slopes of beams for
various loadings and types of support.
– Such a table can be found in the textbook
(Table B19) and provided herein in the
next few viewgraphs (Table 1 and 2).
LECTURE 19. BEAMS: DEFORMATION BY SUPERPOSITION (9.7 – 9.8)
Deflection by Superposition
„
Slide No. 23
ENES 220 ©Assakkaf
Slopes and Deflection Tables
Table 1a
12
LECTURE 19. BEAMS: DEFORMATION BY SUPERPOSITION (9.7 – 9.8)
Deflection by Superposition
„
Slide No. 24
ENES 220 ©Assakkaf
Slopes and Deflection Tables
Table 1b
LECTURE 19. BEAMS: DEFORMATION BY SUPERPOSITION (9.7 – 9.8)
Deflection by Superposition
„
Slide No. 25
ENES 220 ©Assakkaf
Slopes and Deflection Tables
Table 1c
13
LECTURE 19. BEAMS: DEFORMATION BY SUPERPOSITION (9.7 – 9.8)
Deflection by Superposition
„
Slide No. 26
ENES 220 ©Assakkaf
Slopes and Deflection Tables
Table 1d
LECTURE 19. BEAMS: DEFORMATION BY SUPERPOSITION (9.7 – 9.8)
Deflection by Superposition
„
Slide No. 27
ENES 220 ©Assakkaf
Slopes and Deflection Tables
Table 2a
(Beer and Johnston 1992)
14
Slide No. 28
LECTURE 19. BEAMS: DEFORMATION BY SUPERPOSITION (9.7 – 9.8)
Deflection by Superposition
„
ENES 220 ©Assakkaf
Slopes and Deflection Tables
Table 2b
(Beer and Johnston 1992)
LECTURE 19. BEAMS: DEFORMATION BY SUPERPOSITION (9.7 – 9.8)
Deflection by Superposition
„
Slide No. 29
ENES 220 ©Assakkaf
Slopes and Deflection Tables
(Beer and Johnston 1992)
Table 2c
15
Slide No. 30
LECTURE 19. BEAMS: DEFORMATION BY SUPERPOSITION (9.7 – 9.8)
ENES 220 ©Assakkaf
Deflection by Superposition
Use of Slopes and Deflection Tables
„
– Notice that the slope and deflection of the
beam of Figures 21 and 24 (repeated here)
of the illustrative example could have been
determined from the table (Table 1)
y
2m
150 kN
20 kN/m
Figure 21
x
•
D
L=8m
Slide No. 31
LECTURE 19. BEAMS: DEFORMATION BY SUPERPOSITION (9.7 – 9.8)
ENES 220 ©Assakkaf
Deflection by Superposition
„
y
Use of Slopes and Deflection Tables
150 kN
2m
y
20 kN/m
y
2m
x
D
x
D
L=8m
δD
θD
150 kN
=
(θ D )P
L=8m
(δ D )P
+
2m
20 kN/m
x
D
L=8m
(δ D )w
θ D = (θ D ) p + (θ D )w and y D = ( y D ) p + ( y D )w
(θ D )w
Figure 24. Total Slope and Deflection of Point D
16
Slide No. 32
LECTURE 19. BEAMS: DEFORMATION BY SUPERPOSITION (9.7 – 9.8)
Deflection by Superposition
„
ENES 220 ©Assakkaf
Use of Slopes and Deflection Tables
– Indeed, given the information given under
cases 5 and 6 of Tables 2c, the slope and
deflection for any value x ≤ L/4 could have
been expressed analytically.
– Taking the derivative of the expression
obtained in this way, would have yielded
the slope of the beam over the same
interval.
Slide No. 33
LECTURE 19. BEAMS: DEFORMATION BY SUPERPOSITION (9.7 – 9.8)
Deflection by Superposition
„
ENES 220 ©Assakkaf
Use of Slopes and Deflection Tables
– The slope at both ends of the beam may
be obtained by simply adding the
corresponding values given in the table.
– However, the maximum deflection of the
beam of Fig. 21 cannot be obtained by
adding the maximum deflections of cases 5
and 6 (Table 2c), since these deflections
occur at different points of the beam.
17
Slide No. 34
LECTURE 19. BEAMS: DEFORMATION BY SUPERPOSITION (9.7 – 9.8)
ENES 220 ©Assakkaf
Deflection by Superposition
„
Use of Slopes and Deflection Tables
– Applying case 5 on the illustrative example
to find both the slope and deflection of
point D of the beam (Fig. 21), yields
)] (
) ]
) [ (
(θ ) = dy = Pb [3x − (L − b )] = 150 ×10 (6) [3(2) − (8 − 6 )] = −0.003 rad
dx 6 EIL
6(100 × 10 )(8)
( y D )P =
[
Pb 3
150 ×103 (6) 3
2 − 82 − 6 2 (2 ) = −0.009 m
x − L2 − b 2 x =
6 EIL
6 100 × 106 (8)
(
2
2
3
2
D P
2
2
2
6
– These values confirm the results obtained
using Eq. 25 of the integration method.
Slide No. 35
LECTURE 19. BEAMS: DEFORMATION BY SUPERPOSITION (9.7 – 9.8)
ENES 220 ©Assakkaf
Deflection by Superposition
„
Example 6
Use the method of superposition to find the
slope and deflection at point B of the
beam. y
w
C
A
L
2
B
x
L
2
18
Slide No. 36
LECTURE 19. BEAMS: DEFORMATION BY SUPERPOSITION (9.7 – 9.8)
ENES 220 ©Assakkaf
Deflection by Superposition
„
Example 6 (cont’d)
The given loading can be obtained by
superposing the loadings shown in the
following “picture equation” (Fig. 25). The
beam AB is, of course, the same in each
part of the figure.
For each the loadings 1 and 2, the slope
and deflection at B can be determined by
using the Tables 1 or 2. (Textbook Table
B-19)
Slide No. 37
LECTURE 19. BEAMS: DEFORMATION BY SUPERPOSITION (9.7 – 9.8)
ENES 220 ©Assakkaf
Deflection by Superposition
Example 6 (cont’d)
„
C
A
L
2
A
w
B
L
2
=
C
yB
=
θB
Figure 25
Loading 1
C w
A
Loading 2
B
L
2
L
2
C
A
( yB )1
B
w
L
2
(θ B )2
B
(θ B )1
+
C
A
+
A
L
2
B
C
( yB )2
19
Slide No. 38
LECTURE 19. BEAMS: DEFORMATION BY SUPERPOSITION (9.7 – 9.8)
Problem 6 (cont’d)
ENES 220 ©Assakkaf
For the beam and loading shown,
determine the slope and deflection at
point B.
SOLUTION:
Superpose the deformations due to Loading I and Loading II as shown.
Slide No. 39
LECTURE 19. BEAMS: DEFORMATION BY SUPERPOSITION (9.7 – 9.8)
Deflection by Superposition
„
ENES 220 ©Assakkaf
Example 6 (cont’d)
Loading 1:
• From Table 1a or Table 2a (also Table B-19 of
the textbook),
3
(θ B )1 = − wL
6 EI
Loading 2:
4
and
( y B )1 = − wL
8EI
(27a)
• From the same tables:
(θ C )2
w(L / 2)
wL3
=+
=+
6 EI
48 EI
3
and
( yC )2
w(L / 2)
wL4
=+
=
8 EI
128EI
4
(27b)
20
Slide No. 40
LECTURE 19. BEAMS: DEFORMATION BY SUPERPOSITION (9.7 – 9.8)
ENES 220 ©Assakkaf
Deflection by Superposition
„
Example 6 (cont’d)
A
Loading 2
C
w
L
2
(θ B )2
(θ C )2
A
B
Figure 26
In portion CB, the bending moment for loading 2
is zero, thus the elastic curve is a straight line:
wL3
48 EI
(28)
( yB )2 = ( yC )2 + L (θ C )2
(29)
(θ B )2 = (θ C )2 = +
L
2
Slope = (θ C )2
2
( yB )2
B
C
( yC )2
( yC )2
L
2
( yB )2
LECTURE 19. BEAMS: DEFORMATION BY SUPERPOSITION (9.7 – 9.8)
Deflection by Superposition
„
Slide No. 41
ENES 220 ©Assakkaf
Example 6 (cont’d)
Total slope and deflection:
• Slope of Point B:
θ B = (θ B )1 + (θ B )2 = −
wL3 wL3
7 wL3
+
=−
6 EI 48EI
48EI
• Deflection of Point B:
wL4
L  wL3 
7 wL4
 = +
+ 
2
128 EI 2  48EI 
384 EI
wL4 7 wL4
41wL4
y B = ( y B )1 + ( y B )2 = −
+
=−
8 EI 384 EI
384 EI
( y B )2 = ( yC )2 + L (θ C )2 =
21
Slide No. 42
LECTURE 19. BEAMS: DEFORMATION BY SUPERPOSITION (9.7 – 9.8)
ENES 220 ©Assakkaf
Deflection by Superposition
Example 7
„
Use the method of superposition,
determine the deflection at the free end of
the cantilever beam shown in Fig. 27 in
terms of w, L, E, and I.
y
wL
w
Figure 27
A
L
x
C
B
L
Slide No. 43
LECTURE 19. BEAMS: DEFORMATION BY SUPERPOSITION (9.7 – 9.8)
ENES 220 ©Assakkaf
Deflection by Superposition
Example 7 (cont’d)
„
Loading 1
wL
w
A
C
B
L
Figure 28
=
A
C
B
L
L
wL
L
+
w
Loading 2
C
A
B
L
L
Straight Line
A
=
B
δC
θC
B
A
(δ C )1
C
(θ C )1
+
C
A
(δ C )2
B
(θ C )2
22
Slide No. 44
LECTURE 19. BEAMS: DEFORMATION BY SUPERPOSITION (9.7 – 9.8)
Deflection by Superposition
„
ENES 220 ©Assakkaf
Example 7 (cont’d)
Using the solutions listed in Table 1a.
Cases 1 and 2 (Textbook Table B-19) with
P = wL
δ C = (δ C )1 + (δ C )2 = (δ C )1 + (δ B )2 + L(θ B )2
=−
 wL3 
P (2 L)3  wL4

+ −
− L
3EI
8
EI
6
EI



=−
3
 wL3 
71wL4
wL(2 L )  wL4
 = −
+ −
− L
3EI
24 EI
 6 EI 
 8EI
Slide No. 45
LECTURE 19. BEAMS: DEFORMATION BY SUPERPOSITION (9.7 – 9.8)
Deflection by Superposition
„
ENES 220 ©Assakkaf
Example 8
For the simply supported beam of Fig. 29,
use the method of superposition to
determine the total deflection at point C in
terms of P, L, E, and I.
y
Figure 29
A
P
•
B
L/4
L/4
P
•
C
D
x
L
23
Slide No. 46
LECTURE 19. BEAMS: DEFORMATION BY SUPERPOSITION (9.7 – 9.8)
ENES 220 ©Assakkaf
Deflection by Superposition
Example 8 (cont’d)
„
y
A
P
B
y
P
C
• •
L/4 L/4
D
=
x
A
Loading 2
Loading 1
P
B
C
• •
L/4 L/4
L
y
D
x
+
B
x = L/2
ycenter
PL3
=−
48 EI
D
x
L
From Table 1b (Text B-19)
Case 5
a = 3L / 4, b = L / 4
ycenter = −
LECTURE 19. BEAMS: DEFORMATION BY SUPERPOSITION (9.7 – 9.8)
Deflection by Superposition
„
C
• •
L/4 L/4
L
From Table 1c (Text B-19)
Case 6
Figure 30
A
P
Pb(3L2 − 4b 2 )
48 EI
Slide No. 47
ENES 220 ©Assakkaf
Example 8 (cont’d)
Table 1b
24
Slide No. 48
LECTURE 19. BEAMS: DEFORMATION BY SUPERPOSITION (9.7 – 9.8)
Deflection by Superposition
„
Example 8 (cont’d)
ENES 220 ©Assakkaf
Table 1c
Slide No. 49
LECTURE 19. BEAMS: DEFORMATION BY SUPERPOSITION (9.7 – 9.8)
Deflection by Superposition
„
ENES 220 ©Assakkaf
Example 8 (cont’d)
Deflection due to Loading 1:
( yC )1 =
PL3
48 EI
Deflection due to Loading 2:
2
2
2
2
3
( y ) = − Pb(3L − 4b ) = − P(L / 4)[3L − 4(L / 4) ] = − 11PL
C 2
48 EI
48 EI
768 EI
Therefore, total deflection of point C
yC = ( yC )1 + ( yC )2 = −
PL3 11PL3
9 PL3
−
=−
48 EI 768 EI
256 EI
25
Slide No. 50
LECTURE 19. BEAMS: DEFORMATION BY SUPERPOSITION (9.7 – 9.8)
ENES 220 ©Assakkaf
Deflection by Superposition
„
Example 9
Using the method of superposition, find the
deflection at a point midway between the
supports of the beam shown in the figure in
terms of w, L, E, and I.
wL/4
y
w
A
Figure 31
B
L/2
L
x
D
C
3L/4
Slide No. 51
LECTURE 19. BEAMS: DEFORMATION BY SUPERPOSITION (9.7 – 9.8)
ENES 220 ©Assakkaf
Deflection by Superposition
„
Example 9 (cont’d)
The deflection at a point midway between
the supports can be determined by
considering the beam shown in Fig. 32.
Note that since the shear forces VB and VC
do not contribute to the deflection at any
point in span BC, the mid-span deflection
can be expressed as
δ mid = δ M + δ M
B
C
(28)
26
Slide No. 52
LECTURE 19. BEAMS: DEFORMATION BY SUPERPOSITION (9.7 – 9.8)
ENES 220 ©Assakkaf
Deflection by Superposition
Example 9 (cont’d)
„
y
wL/4
Figure 32
w
A
B
L/2
L
yV
MB =
B
wL  L 
 
2 4
=
x
3L/4
wL
2
VC =
B
wL2
=
8
D
C
wL
4
x
C
MC =
L
wL  3L  3wL
 =
4  4  16
Slide No. 53
LECTURE 19. BEAMS: DEFORMATION BY SUPERPOSITION (9.7 – 9.8)
ENES 220 ©Assakkaf
Deflection by Superposition
„
Example 9 (cont’d)
Using the solutions listed in Table 1, Table
2, or Table B-19 of the textbook with
MB = wL2/8 and MC = 3wL2/16
δ mid = δ M + δ M =
B
=
C
(wL / 8)(L ) + (3wL
2
16 EI
2
2
)( )
/ 16 L2
16 EI
4
5wL
256 EI
27
Slide No. 54
LECTURE 19. BEAMS: DEFORMATION BY SUPERPOSITION (9.7 – 9.8)
Deflection by Superposition
„
Example 9 (cont’d)
Table 1d
Slide No. 55
LECTURE 19. BEAMS: DEFORMATION BY SUPERPOSITION (9.7 – 9.8)
Deflection by Superposition
„
ENES 220 ©Assakkaf
ENES 220 ©Assakkaf
Example 10
For the beam in Fig. 33, determine the
flexural stress at point A and the deflection
of the left-hand end.
w = 5 psi
A
6 in
A
Figure 33
4 in
80 in
P = 600 lb
E = 2.4 ×106 psi
28
Slide No. 56
LECTURE 19. BEAMS: DEFORMATION BY SUPERPOSITION (9.7 – 9.8)
ENES 220 ©Assakkaf
Deflection by Superposition
„
Example 10 (cont’d)
The stress at point A is a combination of
compressive flexural stress due to the
concentrated load and a tensile flexural
stress due to the distributed load, hence,
σA =
[
]
M z (3) M y (2 ) 5(4 )(80 ) / 2 (3) 600(80 )(2 )
−
=
−
3
3
Iz
Iy
4(6 ) / 12
6(4 ) / 12
2
= 2,666.7 + 3000.0 = −333.3 psi (compression)
Slide No. 57
LECTURE 19. BEAMS: DEFORMATION BY SUPERPOSITION (9.7 – 9.8)
ENES 220 ©Assakkaf
Deflection by Superposition
„
Example 10 (cont’d)
E = 2.4 ×106 psi
w = 5 psi
A
6 in
80 in
A
4 in
Figure 34
P = 600 lb
δy
δz
y
A
δ = δ y2 + δ z2
z
29
Slide No. 58
LECTURE 19. BEAMS: DEFORMATION BY SUPERPOSITION (9.7 – 9.8)
ENES 220 ©Assakkaf
Deflection by Superposition
„
Example 10 (cont’d)
The deflection at the end of a cantilever
beam with uniformly distributed load is
given by (see Table 1a, case 2)
5(4 )(80)
wL4
=
= 0.5926 in
8EI z 8 2.4 ×10 6 4(6)3 / 12
4
y0 =
(
)[
]
and with concentrated load at the end is
given by (see Table 1a, case 1)
3
PL3
600(80 )
z0 =
=
= 1.3333 in
3EI 3(2.4 ×106 ) [6(4)3 / 12]
LECTURE 19. BEAMS: DEFORMATION BY SUPERPOSITION (9.7 – 9.8)
Deflection by Superposition
„
Example 10 (cont’d)
Slide No. 59
ENES 220 ©Assakkaf
Table 1a
30
LECTURE 19. BEAMS: DEFORMATION BY SUPERPOSITION (9.7 – 9.8)
Deflection by Superposition
„
Slide No. 60
ENES 220 ©Assakkaf
Example 10 (cont’d)
Superimposing the results for the
deflections due to the concentrated and
distributed loads, the deflection at the free
end is the vector sum:
δ = y02 + z02 =
(0.5626)2 + (1.3333)2
= 1.447 in
LECTURE 19. BEAMS: DEFORMATION BY SUPERPOSITION (9.7 – 9.8)
Statically Indeterminate
Transversely Loaded Beams
„
Slide No. 61
ENES 220 ©Assakkaf
The Superposition Method
– The concept of the superposition, which
states that a slope or deflection due to
several loads is the algebraic sum of the
slopes or deflections to each individual
loads acting alone can be applied to
statically indeterminate beams.
– The superposition can provide the
additional equations needed in the
analysis.
31
Slide No. 62
LECTURE 19. BEAMS: DEFORMATION BY SUPERPOSITION (9.7 – 9.8)
Statically Indeterminate
Transversely Loaded Beams
„
ENES 220 ©Assakkaf
The Superposition Method
– Procedure
• Selected restraints are removed and replaced by
unknown loads, e.g., forces and couples.
• Sketching of the deformation (deflection)
diagrams corresponding to individual loads (both
known and unknown).
• Adding up algebraically the individual of
components of slopes or deflections to produce
the known configuration.
Slide No. 63
LECTURE 19. BEAMS: DEFORMATION BY SUPERPOSITION (9.7 – 9.8)
Statically Indeterminate
Transversely Loaded Beams
„
ENES 220 ©Assakkaf
Illustrative Example using Superposition
Determine the reactions at the supports for
the simply supported cantilever beam
(Fig.35) presented earlier for the
integration method.
w
B
A
L
32
Slide No. 64
LECTURE 19. BEAMS: DEFORMATION BY SUPERPOSITION (9.7 – 9.8)
ENES 220 ©Assakkaf
Statically Indeterminate
Transversely Loaded Beams
„
Illustrative Example using Superposition
Method (cont’d)
– First consider the reaction at B as
redundant and release the beam from the
support (remove restraint).
– The reaction RB is now considered as an
unknown load (see Fig. 39) and will be
determined from the condition that the
deflection at B must be zero.
Slide No. 65
LECTURE 19. BEAMS: DEFORMATION BY SUPERPOSITION (9.7 – 9.8)
ENES 220 ©Assakkaf
Statically Indeterminate
Transversely Loaded Beams
„
Illustrative Example using Superposition
Method (cont’d)
w
B
A
A
L
L
RB
(a)
=
B
w
(b)
B
+
( y B )w
A
RB
( y B )R
B
(c)
Figure 39. Original Loading is Broken into Two Loads
33
Slide No. 66
LECTURE 19. BEAMS: DEFORMATION BY SUPERPOSITION (9.7 – 9.8)
Application of Superposition to Statically
Indeterminate Beams
ENES 220 ©Assakkaf
• Determine the beam deformation
• Method of superposition may be
without the redundant support.
applied to determine the reactions at
the supports of statically indeterminate
• Treat the redundant reaction as an
beams.
unknown load which, together with
• Designate one of the reactions as
the other loads, must produce
redundant and eliminate or modify
deformations compatible with the
the support.
original supports.
Slide No. 67
LECTURE 19. BEAMS: DEFORMATION BY SUPERPOSITION (9.7 – 9.8)
ENES 220 ©Assakkaf
Statically Indeterminate
Transversely Loaded Beams
„
Illustrative Example using Superposition
Method (cont’d)
In reference to Table 1a cases 1 and 2
(Table B19 of Textbook):
( y B )R
B
=+
RB L3
3EI
4
and
( y B )w = − wL
8 EI
(37)
The deflection at B in the original structural
configuration must equal to zero, that is
y B = ( y B )RA + ( y B )w = 0
(38)
34
Slide No. 68
LECTURE 19. BEAMS: DEFORMATION BY SUPERPOSITION (9.7 – 9.8)
ENES 220 ©Assakkaf
Statically Indeterminate
Transversely Loaded Beams
„
Slopes and Deflection Tables
Table 1a
Slide No. 69
LECTURE 19. BEAMS: DEFORMATION BY SUPERPOSITION (9.7 – 9.8)
ENES 220 ©Assakkaf
Statically Indeterminate
Transversely Loaded Beams
„
Illustrative Example using Superposition
Method (cont’d)
Substituting Eq. 37 into Eq. 38, gives
RB L3 wL4
−
=0
3EI 8 EI
Solving for RB, the result is
+
3
RB = + wL
8
(39)
(40)
35
Slide No. 70
LECTURE 19. BEAMS: DEFORMATION BY SUPERPOSITION (9.7 – 9.8)
ENES 220 ©Assakkaf
Statically Indeterminate
Transversely Loaded Beams
„
Illustrative Example using Superposition
Method (cont’d)
From the free-body diagram for entire
beam (Figure 40), the equations of
equilibrium are used to find the rest of the
reactions.
+ ↑ ∑ Fy = 0; RAy + RB − wL = 0
(41)
∴ RAy = wL − RB
Slide No. 71
LECTURE 19. BEAMS: DEFORMATION BY SUPERPOSITION (9.7 – 9.8)
Statically Indeterminate
Transversely Loaded Beams
„
ENES 220 ©Assakkaf
Illustrative Example using Superposition
Method (cont’d)
L/2
MA
RAx
wL
w
A
RAy
B
x
RB
Figure 40. Free-body Diagram for the Entire Beam
36
LECTURE 19. BEAMS: DEFORMATION BY SUPERPOSITION (9.7 – 9.8)
Statically Indeterminate
Transversely Loaded Beams
„
Slide No. 72
ENES 220 ©Assakkaf
Illustrative Example using Superposition
Method (cont’d)
3
But RB = wL from Eq. 40, therefore
8
3
5
(42)
RA = wL − wL = wL
8
8
L
+ ∑ M A = 0; - M A − RB L + (wL ) = 0
2
1
1

3
∴ M A = − RB L + wL2 =  wL  L − wL2
2
8
2


1
(43)
= wL2
8
LECTURE 19. BEAMS: DEFORMATION BY SUPERPOSITION (9.7 – 9.8)
Statically Indeterminate
Transversely Loaded Beams
„
Slide No. 73
ENES 220 ©Assakkaf
Illustrative Example using Superposition
Method (cont’d)
From Eqs.40, 42, and 43,
RAx = 0
1
M A = wL2
8
5
RAy = wL
8
3
RB = wL
8
Which confirms the results found by using the
integration method.
37
Slide No. 74
LECTURE 19. BEAMS: DEFORMATION BY SUPERPOSITION (9.7 – 9.8)
ENES 220 ©Assakkaf
Statically Indeterminate
Transversely Loaded Beams
Example 12
„
A beam is loaded and supported as shown
in the figure. Determine (a) the reaction at
supports A and B in terms of w and L, and
(b) the deflection at the left end of the
distributed load in terms of w, L, E, and I.
2wL
w
A
B
C
L
2L
Slide No. 75
LECTURE 19. BEAMS: DEFORMATION BY SUPERPOSITION (9.7 – 9.8)
ENES 220 ©Assakkaf
Statically Indeterminate
Transversely Loaded Beams
Example 12 (cont’d)
„
2wL
w
A
=
B
C
L
2L
w
C
B
A
θ Cw
δ Aw
θ Aw
The portion AC in Figs. 41a and B is a straight line.
θ AR
δ AP
δ Cw
(a)
+
θ AP
Figure 41
δ AR
2wL
A
C
B
δ CR
A
+
A
(b)
RA
A
B
C
θ CR
θ CP
δ CP
A
A
(c)
38
Slide No. 76
LECTURE 19. BEAMS: DEFORMATION BY SUPERPOSITION (9.7 – 9.8)
ENES 220 ©Assakkaf
Statically Indeterminate
Transversely Loaded Beams
„
Example 12 (cont’d)
Note that the portion AC of the beam in
Figs. 41a and 41b is a straight line,
therefore
(a) Using the solution listed in Table 1a
with P = 2wL
δ A = δ Cw + θ Cw (L ) + δ CP + θ CP (L ) + δ AR = 0
(44)
4
3
3
2
3
w(2 L ) w(2 L )
(L ) − 2wL(2 L ) − 2wL(2 L ) (L ) + RA (3L ) = 0
−
−
A
8 EI
6 EI
3EI
2 EI
LECTURE 19. BEAMS: DEFORMATION BY SUPERPOSITION (9.7 – 9.8)
Statically Indeterminate
Transversely Loaded Beams
„
Slopes and Deflection Tables
3EI
Slide No. 77
ENES 220 ©Assakkaf
Table 1a
39
Slide No. 78
LECTURE 19. BEAMS: DEFORMATION BY SUPERPOSITION (9.7 – 9.8)
ENES 220 ©Assakkaf
Statically Indeterminate
Transversely Loaded Beams
„
Example 12 (cont’d)
From which (Eq. 44),
2wL
RA = +
w
C
A
B
L
2L
Equilibrium equations give
MB
RA
38wL
27
+ ↑ ∑ Fy = 0; RA − P − w(2 L ) + RB = 0
RB
FBD
38wL
− 2 wL − 2 wL + RB = 0
3EI
70 wL
∴ RB =
27
Slide No. 79
LECTURE 19. BEAMS: DEFORMATION BY SUPERPOSITION (9.7 – 9.8)
ENES 220 ©Assakkaf
Statically Indeterminate
Transversely Loaded Beams
2wL
„
Example 12 (cont’d)
+
A
B
L
RA
2L
MB
RB
FBD
B
= 0; R A (3L ) − P(2 L ) − w(2 L )(L ) + M B = 0
38wL
(3L ) − 4wL2 − 2wL2 + M B = 0
27
16 wL2
∴MB =
9
w
C
∑M
(b) Deflection at left end of distributed load (at C):
38wL
27
2
38wL
(L ) = 38wL
M C = R A (L ) =
27
27
δ C = δ CRC + δ CM C + δ CP + δ Cw
RC = R A =
(38wL / 27 )(2 L )3 + (38wL2 / 27 )(2 L )2 − 2wL(2 L )3 − w(2 L )4
3EI
2 EI
3EI
8 EI
=−
62 wL4
81EI
40
LECTURE 19. BEAMS: DEFORMATION BY SUPERPOSITION (9.7 – 9.8)
Statically Indeterminate
Transversely Loaded Beams
„
Slopes and Deflection Tables
Slide No. 80
ENES 220 ©Assakkaf
Table 1b
41