1
Deflection of Beams:
1. Equations for Beam Deflection
1.1. Model Elements
• Equilibrium: From the statics of beams,
)()(
)(
2
2
xqdx
Md
xqdx
dV
xVdx
dMz
y
yz
=→
=
= (a)
• The Stress Resultants are obtained from the stresses as
∫∫σ−=A
xxz ydAM (b)
∫∫ τ−=A
xyy dAV (c)
• Stress Strain Relationship: For linearly elastic material
xxxx Eε=σ (d)
• From the Notes on Strains, and Stresses,
2
2
)()( dx
vdyxyxy
xx −≈κ−=ρ−=ε (e)
(where v(x)= transverse deflection of the beam.)
CE 206 - Deflection of Beams 10/17/01
2
1.2. Model Assembly
• Substituting (e) into (d) into (b) and integrating
2
2
)()(
)(dx
vdEIxEIx
EIxM zzzzzz
z ≈κ=ρ
= [1]
• Differentiating eq. [1],
( )xVdx
vdEIdxd
dxdM
yzzz =
= 2
2
(f)
)(2
2
2
2
2
2
xqdx
vdEIdxd
dxMd
dxdV
zzzy =
== [2]
Either [1] or [2] may be used to obtain deflections of beams.
If E, and zzI are constant, the above equations reduce to
zzEIxM
dxvd )(2
2
= [1']
zzEIxV
dxvd )(3
3
= (f')
zzEIxq
dxvd )(4
4
= [2']
CE 206 - Deflection of Beams 10/17/01
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• Equation [1] or [1'] are 2d order o.d.e.s, which requires 2 boundary conditions to yield unique solutions. Equilibrium is imposed separately using eq. (a)
• Equation [2] or [2'] is a 4th order o.d.e., which requires 4 boundary conditions (two at each end) to yield unique solutions.
• Equations (f) or (f') impose only one equilibrium condition. These are typically used for boundary condition formulation only.
1.3. Solution Strategies
• Strategy 1: For a statically determinate beam, we may find )(xM z by statics. Then, the differential equation
zz
zzzz EI
xMdx
vdxMdx
vdEI )()( 2
2
2
2
=→=
may be integrated directly for v', and v(x). (Two boundary conditions needed)
• Strategy 2: The governing differential equation
)(2
2
2
2
xqdx
vdEIdxd
zz =
may be integrated to obtain v, v', v", v''' as functions of x. (4 boundary conditions are needed.)
CE 206 - Deflection of Beams 10/17/01
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1.4. Boundary Conditions: Depending upon whether 2d order or 4th order equations are to be solved, we have several possible boundary conditions to be considered. • If moment is imposed on the end, note that
zz
z
EIM
dxvd =2
2
• If shear is imposed on the end, note that
yzz Vdx
vdEIdxd =
2
2
These ideas allow a variety of boundary conditions to be handled. Also note that • For 2d order o.d.e's, a total of 2 b.c. are needed. • For 4th order o.d.e.'s, a total or 4 b.c. are needed.
CE 206 - Deflection of Beams 10/17/01
5
Boundary Conditions to be applied:
1. All 2d order BC also are used for 4th order o.d.e. 2. 4th order BC are established by statics for 2d order o.d.e. Condition Appearance 2d Order
BC 4th Order BC
Fixed end 0 ,0 ==dxdvv
same
Simple support 0=v 0''
0=→
=v
M z
Free end See note 3. 0)'''(
0''==
==vEIVvEIM
zzy
zzz
intermediate roller
'' 21
21
vvvv
==
'''' 21 21
21
vEIvEIMM
zzzz
zz
=→
=
Concentrated end load
See note 3
0'')'''(
0
===→
=
vEIMPvEI
PV
zzz
ozz
y
Concentrated end moment
See note 3. 0''
0)'''(MvEIM
vEIV
zzz
zzy
−==
==
Concentrated intermediate load
'' 21
21
vvvv
==
02
21
)'''()'''(
''''21
PvEIvEIvEIvEI
zzzz
zzzz
=−
=
Concentrated intermediate moment
'' 21
21
vvvv
==
012
21
''''
)'''()'''(21
MvEIvEIvEIvEI
zzzz
zzzz
−=−
=
Intermediate hinge
21 vv = (no slope continuity)
0'' ,0''
)'''()'''(
21
012 12
==
=−
vvPvEIvEI zzzz
CE 206 - Deflection of Beams 10/17/01
6
2. Slope and Deflection by Integration
2.1. Statically Determinate Beams
For these beams, it's often convenient to do the statics separately and use the 2d order o.d.e.'s... S.P. 1: Uniformly loaded cantilever beam
( ) 2/2)(
2/))(()(022
0
0
xLxLwxM
xLxLwxMM
z
zz
−−−=→
−−−−==∑
Therefore,
2/)2('' 220 xLxLwvEI zz +−−=
Subject to the B.C.
v(0)=0, v'(0)=0. We may solve by either direct or indirect integration, since there are no lower degree derivatives in v in the o.d.e.
CE 206 - Deflection of Beams 10/17/01
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Direct Integration Solution:
(i) First integration:
)
3(
2)('
)3
(2
)0(')('
3220
3220
xLxxLEIwxv
xLxxLEIwvxv
zz
zz
+−−
=→
+−−
=−
(ii) Second integration:
+−−
=→
+−−
=−
12322)(
12322)0()(
43220
43220
xLxxLEIwxv
xLxxLEIwvxv
zz
zz
Indirect integration solution: (i) Integrate twice, picking up constants of integration along the way.
21
43220
1
3220
12322)(
32'
CxCxLxxLEIwxv
CxLxxLEIwv
zz
zz
++
+−−=
+
+−−
=
CE 206 - Deflection of Beams 10/17/01
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(ii) Substitute the boundary conditions at x=0:
0)0('0)0(
1
2
====
CvCv
Therefore,
+−−=12322
)(4322
0 xLxxLEIwxv
zz
the same as was obtained by direct integration.
It is informative to write this solution as
+
−
−
=4324
0
31
342
8)(
Lx
Lx
Lx
EILwxvzz
Note that the displacement caused by the uniform load is proportional to the ratio zzEILw /4
0 , which has units of displacement. at x=L. the tip deflection is
zzEILwLv
8)(
40−
=
CE 206 - Deflection of Beams 10/17/01
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S.P. 2: Solution of the 4th order o.d.e.
(This solution incorporates the statics into the d.e. )
The equation is
oiv
zz wvEI −=
subject to the boundary conditions
shear) (zero 0)('''moment) (zero 0)(''
0)0('0)0(
==
==
LvLv
vv
Indirect integration:
432231
40
3221
30
21
20
10
2624
26'
2''
'''
CxCxCxCEI
xwv
CxCxCEI
xwv
CxCEI
xwv
CEI
xwv
zz
zz
zz
zz
++++−
=
+++−
=
++−
=
+−
=
CE 206 - Deflection of Beams 10/17/01
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The boundary conditions determine the four constants. At x=0:
0)0('0)0(
3
4
====
CvCv
So, the last two constants vanish because of the fixed end at x=0, leaving the solution in the form .
22314
0
2624)( xCxC
EIxwxv
zz
++−
=
At x=L:
zzzz
zz
EILwCC
EILwLv
CLCEI
LwLv
011
0
21
20
0)('''
02
)(''
=→=+−
=
=++−=
Backsubstituting,
zzEI
LwC2
20
2 −=
Inserting these constants into the solution gives
zzzzzz EIxLw
EILxw
EIxwxv
4624)(
20
30
40 −+
−=
which is the same as the previous solution.
CE 206 - Deflection of Beams 10/17/01
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S.P. 3: Simply supported beam under end moment
(a) 2d Order Eqn:
By statics, the moment can be shown to be
LxMxM /)( 0=
Therefore:
21
30
1
20
0
6)(
2)('
)(''
CxCLEI
xMxv
CLEI
xMxv
LEIxMxv
zz
zz
zz
++=
+=
=
B.C: 0 0)0( 22 =→== CCv
zzzz EILMCLC
EILMLv
660)( 0
11
20 −=→+==
Solution:
−
=−=
Lx
Lx
EILM
EILxM
LEIxMxv
zzzzzz
3200
30
666)(
CE 206 - Deflection of Beams 10/17/01
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(b) 4th order eqn: No load on the beam, so
43
2
2
3
1
32
2
1
21
1
26
2'
'''''
00
CxCxCxCv
CxCxCv
CxCvCv
vvEI ivivzz
+++=
++=
+==
=→=
B.C:
zz
zzzz
EILMC
LEIMCLC
EIMLv
LCCLCLCLv
CvCv
6
)(''
6
60)(
0C 0)0(''0C 0)0(
03
011
0
2
133
3
1
22
44
−=
=→==
−=→+==
=→===→==
−
=−=
Lx
Lx
EILM
EILxM
LEIxMxv
zzzzzz
3200
30
666)(
as before.
CE 206 - Deflection of Beams 10/17/01
13
S.P. 4: Simply supported beam under sinusoidal load
D.E.: Use
π−=
LxwvEI zz sin')'''( 0
B.C.: 0)( 0)0( == Lvv 0)('' 0)0('' == Lvv
Indirect Integration:
43
23
14
40
32
2
13
30
212
20
10
26sin
2cos'
sin''
cos)'''(
CxCxCxCLx
EILwv
CxCxCLx
EILwv
CxCLxLwvEI
CLxLwvEI
xzz
zz
zz
zz
++++
π
π−=
+++
π
π−=
++
π
π=
+
π
π=
B.C. at x=0:
0)0( 4 == Cv 04 =C
0)0('' 2 == Cv 02 =C
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So the solution reduces to
xCxCLx
EILwv
zz3
3
14
40
6sin ++
π
π−= \
B.C. at x=L:
LCLCLv 3
3
1 60)( +== (Since sin(π)=0)
0)('' 1 == CEI
LLvzz
→ 0 ,0 31 == CC
So, in this case, all constants are zero, and
π
π=
Lx
EILwv sin4
40
The moments and shears are
π
π==
π
π==
LxLwvEIV
LxLwvEIM
zzz
zzz
cos'''
sin''
0
2
20
Note: It isn't obvious here, but this example illustrates an important building block of beam and plate analysis.
CE 206 - Deflection of Beams 10/17/01
15
If the loading is described piecewise on the member or structure, the solution must be constructed by piecewise solutions, with • Boundary conditions at supports or ends • continuity conditions at the points of discontinuous
load description Solution Strategy: (1) Subdivide the beam into sections at the points of
load discontinuity. (2) Solve the o.d.e (either 2d order or 4th order) for
each section. (3) Match continuity conditions between portions and
boundary conditions at the ends. Side Note: In later courses, you'll study much more
efficient ways of doing this. This semester we'll just learn the basic concepts.
CE 206 - Deflection of Beams 10/17/01
16
S.P. 5: Solve using the second order o.d.e.'s (Do the statics separately.)
(a) Beam reactions:
4
42210 00
0LwVLwVLwVF yyyy =→−=−==∑
The resultant of the distributed load acts at 5L/6, so
245
06
54
200 LwMLLwMM zAz −=→=⋅−−=∑
(b) For x < L/2: The beam moment is
424
5
0424
5
02
0
02
0)(
LxwLwM
LxwLwMM
z
zx
z
+−=→
=−+=∑
CE 206 - Deflection of Beams 10/17/01
17
(c) Integrate the moment for 0 < x < L/2:
21302
20
1
120
20
1
02
01
24485
8245'
4245)(''
CxCxEI
LwxEI
Lwv
CxEI
LwxEI
Lwv
EILxw
EILw
EIxMv
zzzz
zzzz
zzzzzz
+++−=
++−=
+−==
From the cantilever beam boundary conditions at x=0:
0)0('0)0(
11
21
====
CvCv
So, the solution for 0 < x < L/2 reduces to
3022
01 2448
5 xEI
LwxEI
Lwvzzzz
+−=
This solution only applies over the first half of the beam. For the second half, the distributed load must be properly accounted for.
CE 206 - Deflection of Beams 10/17/01
18
(d) Determine the moment for x < L/2:
LxwxwLwM
LxL
wxLwLwM
LxLxLxw
xLwLwMM
z
z
zx
z
326
0224424
5
231
212
21
4245
30
20
20
300
20
0
02
0)(
−+−=→
=
−+−+=
−
−
−+
−+=∑
(e) Integrate the moment for L/2 < x < L.
43
50
402
20
2
3
40
30
20
2
30
20
20
2
602412
1266'
326''
CxCEIxw
EIxwx
EILwv
CEIxw
EIxwx
EILwv
LEIxw
EIxw
EILw
EIMv
zzzzzz
zzzzzz
zzzzzzzz
z
++−+−=
+−+−=
−+−==
CE 206 - Deflection of Beams 10/17/01
19
(f) To determine 43 and CC , the continuity conditions at L/2 are applied...
)2/(')2/(' )2/()2/( 2121 LvLvLvLv ==
where
zz
zz
EILwLv
EILwLv
967)2/('
48)2/(
30
1
40
1
−=
−=
1
30
2
21
40
2
19213
)2/(
21603
)2/(
CEI
LwLv
CLCEILw
Lv
zz
zz
+−=
++−=
Setting )2/(')2/(' 21 LvLv = :
zzzzzz EILwCC
EILw
EILw
192
19213
967 3
011
30
30 −=→+−=−
Setting )2/()2/( 21 LvLv = : (and substituting 1C from above)
zzzzzz EILwCC
EILw
EILw
1920
192041
48
40
22
40
40 =→+−=−
Therefore, for x > L/2:
zzzzzzzz EILw
xEILw
LEIxw
EIxw
EIxLw
xv1920192602412
)(4
03
05
04
022
02 +−−+−=
CE 206 - Deflection of Beams 10/17/01
20
S.P. 6: We may also use the 4th order o.d.e.'s
Here there's no load except at L/2. The reactions at each of the two ends are P/2. (But we don't need that for the solution.) Assume that zzIE , are constant
To the left of the concentrated load,
432231
1
3221
1
211
11
1
26
2'
'''''
00')'''(
CxCxCxCv
CxCxCv
CxCvCv
vvEI ivzz
+++=
++=
+==
=→=
B.C.: At x=0,
0 0)0(''0 0)0(
221
441
=→===→==
CCvCCv
so,
xCxCv 331
1 6+=
CE 206 - Deflection of Beams 10/17/01
21
To the right of the concentrated load:
87
26352
7625
2
652
52
22
26
2'
'''''
00')'''(
CxC
xCxCv
CxCxCv
CxCvCv
vvEI ivzz
++
+=
++=
+==
=→=
B.C. At x=L,
LCLCCLC
CLCLv
CLCLCLCLv
7
2
5856
652
87
2
6
3
52
2 ,C
0)(''26
0)(
−=−=→
+==
+++==
so
)(326 7
323
52 LxCLxLxCv −+
+−=
We still have 4 undetermined constants, and haven't included the load effect. To take care of this, we need the continuity conditions at the load point.
CE 206 - Deflection of Beams 10/17/01
22
At the load point,
(i) Continuity of displacements:
75
3
31
3
21 24811
24822CLCLCLCLLvLv −=+→
=
(ii) Continuity of slopes:
75
2
31
2
21 83
82'
2' CCLCCLLvLv +−=+→
=
(iii) Continuity of moments:
)2/('')2/('')2/('')2/(''
)2/()2/(
21
21
21
LvLvLEIvLvEI
LMLM
zz
=→=→
=
since zzIE , are constant. Therefore,
515121 22)2/('')2/('' CCLCLCLvLv −=→−=→=
CE 206 - Deflection of Beams 10/17/01
23
(iii) Shear equilibrium:
From equilibrium, at a point load, -P,
PLVLVV yyy −=−=∆ )2/()2/(12
But,
52
11
)'''(
)'''(
2
1
CEIvEIVCEIvEIV
zzzzy
zzzzy
==
==
so
PCEICEI zzzz −=− 15 Using these results together, obtain
zzzz EI
PCEIPC
2 ,
2 51 −==
Substituting 51 and CC into the displacement and slope continuity conditions,
7
3
3
3
224811
2248CL
EIPLCL
EIPL
zzzz
−
−=+
7
2
3
2
283
28C
EIPLC
EIPL
zzzz
+
−−=+
zzEI
PLDPLC163
,16
2
3
2
3 −=−=→
CE 206 - Deflection of Beams 10/17/01
24
Substituting all of these constants into the solutions,
)(
163
3262
16122323
2
23
1
LxEIPLLxLx
EIPv
xEI
PLxEIPv
zzzz
zzzz
−−
+−−=
−=
The displacement at any point of the beam is given by
≤<≤≤
=LxLxv
Lxxvxv
2/ )(2/0 )(
)(2
1
Maximum displacement occurs at L/2, where
zzEI
PLLv48
)2/(3
−=
The - sign means that the displacement is downward, consistent with the downward loading.
CE 206 - Deflection of Beams 10/17/01
25
2.2. Statically Indeterminate Beams
If a beam has more reactions than can be determined by statics, the moments can't be found ahead of time.
The deflected shape of the beam, taking into account the boundary conditions determines the moments. Three strategies are possible:
(a) Perform the statics separately, leaving the excess reactions as unknowns. Then integrate the 2d order o.d.e. and impose the displacement and rotation b.c. to solve for the reactions.
(b) Integrate the 4th order o.d.e. directly, and impose all boundary conditions.
(c) Superimpose solutions of simpler problems to satisfy the boundary conditions.
CE 206 - Deflection of Beams 10/17/01
26
S.P. 7: Analyze the statically indeterminate beam shown.
(a) Conduct the static analysis separately, and incorporate the moments as unknowns:
)(12
2)(1
2/
02)(
0
zBzAL
LowyB
LwBAL
A
LwBALAM
LwBAF
ozzy
ozzyB
Z
yyy
−+=
+−−=→
++−−=
=−+=
∑∑
Note that yy BA and have two components: (i) Effect of load if zz BA , are zero. (ii) Effect of zz BA , .
CE 206 - Deflection of Beams 10/17/01
27
Next, cut a FBD at x and obtain the moment.
LxB
LxAxwLxwxM
xwxBAL
LwAxMM
zzoo
z
ozz
ozz
xZ
+
−+−=→
=
+
−−−−=∑
122
)(
0 2
)(12
)(
2
2)(
Therefore, the curvature is given by
+
−+−=
=
LxB
LxAxwLxw
EI
EIxM
dxvd
zzo
zz
zz
z
122
1
)(
20
2
2
Integrate twice. Before doing this, note that the boundary conditions are
0)(')()0(')0( ==== LvLvvv Here, let's use indirect integration.
CE 206 - Deflection of Beams 10/17/01
28
The first integration yields
1
2320
2641 C
LxB
LxxAxwLxw
EIdxdv
zzo
zz
+
+
−+−=
But, v'(0)=0 implies that ,01 =C so this reduces to
+
−+−=LxB
LxxAxwLxw
EIdxdv
zzo
zz 2641 232
0
The second integration yields
2
232430
26224121)( C
LxB
LxxAxwLxw
EIxv zz
o
zz
+
+
−+−=
But, v(0)=0 implies that 02 =C . So, the displacement is
+
−+−=L
xBL
xxAxwLxw
EIxv zz
o
zz 26224121)(
232430
At this point the boundary conditions at x=0 have been satisfied.
CE 206 - Deflection of Beams 10/17/01
29
We still don't know zz BA and , but we still have the conditions .0)(' ,0)( == LvLv Hence,
02212
)('
06324
)(
3
224
=++=
=++=
zz
z
zz
z
zz
o
zz
z
zz
z
zz
o
EILB
EILA
EILwLv
EILB
EILA
EILwLv
Solving these two equations simultaneously yields
12
20 LwBA zz −==
Therefore, the moments are
1222)(
22 LwxwLxwxM oooz −−=
and the displacements are
zz
o
zz
o
zz EIxLw
EIxw
EILxwxv
242412)(
22430 −−=
These functions are plotted below.
CE 206 - Deflection of Beams 10/17/01
30
Moments as a function of x:
Displacements as a function of x:
Note that • the displacements are zero at the ends. • the slopes are also zero at the ends (as they must be
for the fixed boundary conditions to be satisfied).
CE 206 - Deflection of Beams 10/17/01
31
S.P. 8: Solve the problem of S.P. 7 by integrating the 4th order differential equation
zz
oivo
ivzz EI
wvwvEI −=→−=
First Integration:
1''' CEI
xwvzz
o +−=
Second Integration:
21
2
2'' CxC
EIxwv
zz
o ++−=
Third Integration:
32
2
1
3
26' CxCxC
EIxwv
zz
o +++−=
Fourth Integration:
43
2
2
3
1
4
2624CxCxCxC
EIxwv
zz
o ++++−=
4 constants of integration must be determined from the boundary conditions.
CE 206 - Deflection of Beams 10/17/01
32
BC at x=0:
0 0)0('0 0)0(
33
44
=→===→==
CCvCCv
So, the solution reduces (at this point to
2624
2
2
3
1
4 xCxCEIxwv
zz
o ++−=
BC at x=L:
LCLCEI
LwLv
LCLCEILwLv
zz
o
zz
o
2
2
1
3
2
2
3
1
4
260)('
26240)(
++−==
++−==
Solving these two equations for 21 and CC ,
zz
o
zz
o
EILwC
EILwC
12 ,
2
2
21 −==
The solution then becomes
zz
o
zz
o
zz EIxLw
EIxw
EILxwxv
242412)(
22430 −−=
in agreement with the solution of S.P. 7.
CE 206 - Deflection of Beams 10/17/01
33
This approach can also be used for somewhat more general loads and boundary conditions, although the amount of effort to obtain solutions increases. S.P. 9:
We'll solve this one by integrating the two ends separately and matching boundary conditions. The integrals are the same as in S.P. 6, but the boundary conditions are different. To the left of the load, To the left of the concentrated load,
432231
1
3221
1
211
11
1
26
2'
'''''
00')'''(
CxCxCxCv
CxCxCv
CxCvCv
vvEI ivzz
+++=
++=
+==
=→=
CE 206 - Deflection of Beams 10/17/01
34
B.C.: At x=0,
0 0)0(''0 0)0(
221
441
=→===→==
CCvCCv
so,
xCxCv 331
1 6+=
To the right of the concentrated load:
872635
2
7625
2
652
52
22
26
2'
'''''
00')'''(
CxCxCxCv
CxCxCv
CxCvCv
vvEI ivzz
+++=
++=
+==
=→=
B.C. At x=L,
76
2
52
87
2
6
3
52
20)('
260)(
CLCLCLv
CLCLCLCLv
++==
+++==
(These won't be quite as easy to solve as in S.P. 6, so we'll try a more formal approach.)
CE 206 - Deflection of Beams 10/17/01
35
The continuity conditions at the load point are the same as in S.P. 6. Namely, (i) Continuity of displacements:
87
2
6
3
53
3
1
21
2848248
)2/()2/(
CLCLCLCLCLC
LvLv
+++=+→
=
(ii) Continuity of rotations:
76
2
53
2
1
21
288
)2/(')2/('
CLCLCCLC
LvLv
++=+→
=
(iii) Continuity of moments:
651
21
22
)2/('')2/(''
CLCLC
LvLv
+=→
=
(iv) Shear equilibrium:
zzzz EIPCCPvvEI /)''''''( 5121 =−→=− To solve, let's organize these algebraic boundary and continuity condition equations into a matrix format..
CE 206 - Deflection of Beams 10/17/01
36
=
−
−−
−−−
−−−−
zzEIP
CCCCCC
LL
LLL
LLLLL
LL
LLL
00000
000101
0012
02
0128
18
12848248
012
00
126
00
8
7
6
5
2
1
22
233
2
23
These may either be solved by hand (with considerable effort) or symbolically by computer. (Some possibilities include Mathematica, Mathcad, or Maple.) Using the latter approach,
zzEIP
LL
L
L
CCCCCC
−
−−
=
48/32/5
2/16/1132/
16/5
3
2
2
8
7
6
5
2
1
CE 206 - Deflection of Beams 10/17/01
37
Then, the solution is
LxLEI
PLxLLxxv
LxEI
PxLxv
zz
zz
≤<
+−+−=
≤≤
−=
2
48325
49611
2/0 3296
5
3223
2
23
1
CE 206 - Deflection of Beams 10/17/01
38
3. Analysis of Beam Deflections and Statically Indeterminate Beams by Superposition Key Observation: Because the differential equation for the model is linear, we may superposition (add) solutions to obtain combined solutions!