Deflection of Beams:
1. Equations for Beam Deflection
1.1. Model Elements
• Equilibrium: From the statics of beams,
dM z

= V y ( x)
2
 d Mz
dx
= q( x )
→
2
dV y
dx
= q( x ) 

dx
(a)
• The Stress Resultants are obtained from the
stresses as
M z = − ∫∫ σ xx ydA
(b)
A
V y = ∫∫ − τ xy dA
(c)
A
• Stress Strain Relationship: For linearly elastic
material
σ xx = Eε xx
(d)
• From the Notes on Strains, and Stresses,
ε xx
d 2v
−y
= − yκ ( x ) ≈ − y 2
=
ρ( x )
dx
(where v(x)= transverse deflection of the beam.)
1
(e)
CE 206 - Deflection of Beams
10/17/01
1.2. Model Assembly
• Substituting (e) into (d) into (b) and integrating
EI zz
d 2v
M z ( x) =
= EI zz κ ( x ) ≈ EI zz 2
ρ( x )
dx
[1]
• Differentiating eq. [1],
dM z d 
d 2v 
=  EI zz 2  = V y ( x )
dx
dx 
dx 
d 2M z d 2 
d 2v 
=
= 2  EI zz 2  = q( x )
2
dx
dx
dx 
dx 
dV y
(f)
[2]
Either [1] or [2] may be used to obtain deflections of
beams.
If E, and I zz are constant, the above equations reduce
to
d 2 v M ( x)
=
[1']
2
EI zz
dx
d 3v V ( x)
=
(f')
3
EI zz
dx
d 4 v q( x )
=
[2']
4
EI zz
dx
2
CE 206 - Deflection of Beams
10/17/01
• 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 M z ( x ) by statics. Then, the differential
equation
d 2v
d 2 v M z ( x)
EI zz 2 = M z ( x ) → 2 =
EI zz
dx
dx
may be integrated directly for v', and v(x). (Two
boundary conditions needed)
• Strategy 2: The governing differential equation
d2 
d 2v 
 EI zz 2  = q( x )
2
dx 
dx 
may be integrated to obtain v, v', v", v''' as functions
of x. (4 boundary conditions are needed.)
3
CE 206 - Deflection of Beams
10/17/01
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
d 2v M z
=
2
EI zz
dx
• If shear is imposed on the end, note that
d 
d 2v 
 EI zz 2  = V y
dx 
dx 
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.
4
CE 206 - Deflection of Beams
10/17/01
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
Fixed end
Appearance 2d Order 4th Order BC
BC
v = 0,
dv
=0
dx
same
Mz = 0
Simple support
v=0
Free end
→ v' ' = 0
See note 3. M z = EI zz v ' ' = 0
V y = ( EI zz v ' ' )' = 0
intermediate
roller
Concentrated
end load
v1 = v2
M z1 = M z 2
v1 ' = v2 '
→ EI zz1 v1 ' ' = EI zz 2 v2 ' '
See note 3
V y = P0
→ ( EI zz v ' ' )' = Po
Concentrated
end moment
See note 3.
M z = EI zz v ' ' = 0
V y = ( EI zz v ' ' )' = 0
Concentrated
intermediate
load
Concentrated
intermediate
moment
Intermediate
hinge
v1 = v2
M z = EI zz v ' ' = − M 0
EI zz1 v1 ' ' = EI zz 2 v2 ' '
v1 ' = v2 '
( EI zz v2 ' ' )'−( EI zz v ' ' )' = P0
v1 = v2
( EI zz1 v1 ' ' )' = ( EI zz 2 v2 ' ' )'
v1 ' = v2 '
EI zz v2 ' '− EI zz v1 ' ' = − M 0
v1 = v2
( EI zz 2 v2 ' ' )'−( EI zz1 v1 ' ' )' = P0
(no slope
continuity)
v1 ' ' = 0, v2 ' ' = 0
5
CE 206 - Deflection of Beams
10/17/01
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
∑M
z
= 0 = − M z ( x ) − w0 ( L − x )( L − x ) / 2
→ M z ( x ) = − w0 (L2 − 2 Lx − x 2 ) / 2
Therefore,
EI zz v ' ' = − w0 ( L2 − 2 Lx + x 2 ) / 2
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.
6
CE 206 - Deflection of Beams
10/17/01
Direct Integration Solution:
(i) First integration:
3
− w0 2
x
( L x − Lx 2 + )
v ' ( x ) − v ' ( 0) =
2 EI zz
3
→
− w0 2
x3
2
( L x − Lx + )
v' ( x ) =
2 EI zz
3
(ii) Second integration:
− w0  L2 x 2 Lx 3 x 4 
−
+ 
v ( x ) − v ( 0) =

2 EI zz  2
3 12 
− w0  L2 x 2 Lx 3 x 4 
→
−
+ 
v( x ) =

2 EI zz  2
3 12 
Indirect integration solution:
(i) Integrate twice, picking up constants of integration
along the way.
− w0  2
x3 
2
v' =
 L x − Lx +  + C1
2 EI zz 
3
w0  L2 x 2 Lx 3 x 4 
−
+  + C1 x + C 2
v( x ) = −

2 EI zz  2
3
12 
7
CE 206 - Deflection of Beams
10/17/01
(ii) Substitute the boundary conditions at x=0:
v ( 0) = C 2 = 0
v ' (0) = C1 = 0
Therefore,
w0  L2 x 2 Lx 3 x 4 
−
+ 
v( x ) = −

2 EI zz  2
3
12 
the same as was obtained by direct integration.
It is informative to write this solution as
− w0 L4
v( x ) =
8EI zz
  x 2 4  x 3 1  x 4 
 2  −   +   
3  L 3 L 
  L
Note that the displacement caused by the uniform
load is proportional to the ratio w0 L4 / EI zz , which has
units of displacement.
at x=L. the tip deflection is
− w0 L4
v ( L) =
8EI zz
8
CE 206 - Deflection of Beams
10/17/01
S.P. 2: Solution of the 4th order o.d.e.
(This solution incorporates the statics into the d.e. )
The equation is
EI zz v iv = − wo
subject to the boundary conditions
v ( 0) = 0
v ' ( 0) = 0
v ' ' ( L) = 0 (zero moment)
v ' ' ' ( L) = 0 (zero shear)
Indirect integration:
v' ' ' =
− w0 x
+ C1
EI zz
− w0 x 2
+ C1 x + C 2
v' ' =
2 EI zz
− w0 x 3 C1 2
+
v' =
x + C2 x + C3
6 EI zz
2
− w0 x 4 C1 3 C 2 2
+
v=
x +
x + C3 x + C4
24 EI zz
6
2
9
CE 206 - Deflection of Beams
10/17/01
The boundary conditions determine the four
constants. At x=0:
v ( 0) = C 4 = 0
v ' ( 0) = C 3 = 0
So, the last two constants vanish because of the fixed
end at x=0, leaving the solution in the form .
− w0 x 4 C1 3 C 2 2
+
v( x ) =
x +
x
24 EI zz
6
2
At x=L:
w0 L2
v ' ' ( L) = −
+ C1 L + C 2 = 0
2 EI zz
w0 L
− w0 L
v ' ' ' ( L) =
+ C1 = 0 → C1 =
EI zz
EI zz
Backsubstituting,
w0 L2
C2 = −
2 EI zz
Inserting these constants into the solution gives
− w0 x 4 w0 Lx 3 w0 L2 x
v( x ) =
+
−
24 EI zz 6 EI zz
4 EI zz
which is the same as the previous solution.
10
CE 206 - Deflection of Beams
10/17/01
S.P. 3: Simply supported beam under end moment
(a) 2d Order Eqn:
By statics, the
moment can be shown
to be
M ( x) = M 0 x / L
Therefore:
v' ' ( x ) =
M0x
EI zz L
M 0x2
v' ( x ) =
+ C1
2 EI zz L
M 0x3
v( x ) =
+ C1 x + C 2
6 EI zz L
B.C:
v ( 0) = 0 = C 2
→
C2 = 0
M 0 L2
M 0L
v ( L) = 0 =
+ C1 L → C1 = −
6 EI zz
6 EI zz
Solution:
M 0 x 3 M 0 Lx M 0 L2
v( x ) =
−
=
6 EI zz L 6 EI zz 6 EI zz
11
 x  3 x 
  − 
 L  L 
CE 206 - Deflection of Beams
10/17/01
(b) 4th order eqn: No load on the beam, so
EI zz v iv = 0 → v iv = 0
v ' ' ' = C1
v ' ' = C1 x + C 2
x2
v ' = C1
+ C2 x + C3
2
x3
x2
v = C1
+ C2
+ C3 x + C4
6
2
B.C:
v ( 0) = 0 = C 4
→
C4 = 0
v ' ' ( 0) = 0 = C 2
→
C2 = 0
L3
v ( L) = 0 = C1 + C 3 L →
6
M0
v ' ' ( L) =
= C1 L
→
EI zz
L2
C 3 = −C1
6
M0
C1 =
EI zz L
C3 = −
M 0 x 3 M 0 Lx M 0 L2
v( x ) =
−
=
6 EI zz L 6 EI zz 6 EI zz
as before.
12
M0L
6 EI zz
 x  3 x 
  − 
 L  L 
CE 206 - Deflection of Beams
10/17/01
S.P. 4: Simply supported beam under sinusoidal load
 πx 
D.E.: Use ( EI zz v ' ' )' ' = − w0 sin 
 L
B.C.:
v ( 0) = 0
v ( L) = 0
v ' ' ( 0) = 0
v ' ' ( L) = 0
Indirect Integration:
( EI zz v ' ' )' =
w0 L  πx 
cos  + C1
π
 L
w0 L2  πx 
EI zz v ' ' = 2 sin  + C1 x + C 2
π
 L
w0 L3
x2
 πx 
v' = − 3
cos  + C1
+ C2 x + C3
2
π EI zz
 L
w0 L4
x3
x2
 πx 
v=− 4
sin  + C1
+ Cx
+ C3 x + C4
L
6
2
π EI zz  
B.C. at x=0:
v ( 0) = C 4 = 0
C4 = 0
v ' ' ( 0) = C 2 = 0
C2 = 0
13
CE 206 - Deflection of Beams
10/17/01
So the solution reduces to
w0 L4
x3
 πx 
sin  + C1
+ C3 x \
v=− 4
6
π EI zz  L 
B.C. at x=L:
L3
v ( L) = 0 = C1 + C 3 L
6
L
v ' ' ( L) =
C1 = 0
EI zz
(Since sin(π)=0)
→
C1 = 0, C 3 = 0
So, in this case, all constants are zero, and
w0 L4
 πx 
v = 4 sin 
π EI  L 
The moments and shears are
w0 L2  πx 
M z = EI zz v ' ' = 2 sin 
π
 L
w 0 L  πx 
cos 
V z = EI zz v ' ' ' =
π
 L
Note: It isn't obvious here, but this example illustrates
an important building block of beam and plate
analysis.
14
CE 206 - Deflection of Beams
10/17/01
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.
15
CE 206 - Deflection of Beams
10/17/01
S.P. 5: Solve using the second order o.d.e.'s (Do the
statics separately.)
(a) Beam reactions:
w0 L
w0 L
1 L
F
V
w
V
V
0
=
=
−
=
−
→
=
∑ y
y
y
y
0
2
2
4
4
The resultant of the distributed load acts at 5L/6, so
5w0 L2
w0 L 5 L
∑ M Az = − M − 4 ⋅ 6 = 0 → M z = − 24
(b) For x < L/2: The beam moment is
5w0 L2 w0 Lx
( x)
∑ M z = M z + 24 − 4 = 0
5w0 L2 w0 Lx
→ Mz = −
+
24
4
16
CE 206 - Deflection of Beams
10/17/01
(c) Integrate the moment for 0 < x < L/2:
5w0 L2 w0 Lx
M ( x)
v1 ' ' =
=−
+
EI zz
24 EI zz 4 EI zz
5w0 L2
wL
v1 ' = −
x + 0 x 2 + C1
24 EI zz
8EI zz
5w0 L2 2
wL
v1 = −
x + 0 x 3 + C1 x + C 2
48EI zz
24 EI zz
From the cantilever beam boundary conditions at
x=0:
v1 (0) = C 2 = 0
v1 ' (0) = C1 = 0
So, the solution for 0 < x < L/2 reduces to
5w0 L2 2
wL
v1 = −
x + 0 x3
48EI zz
24 EI zz
This solution only applies over the first half of the
beam. For the second half, the distributed load must
be properly accounted for.
17
CE 206 - Deflection of Beams
10/17/01
(d) Determine the moment for x < L/2:
∑M
( x)
z
5w0 L2 w0 L
x
= Mz +
−
24
4
L1
L
1  2 x 
+ w0  − 1  x −   x − 
2  L
2  3
2

3
w 
5w0 L2 w0 L
L
x + 0 x −  = 0
= Mz +
−
24
4
24 L 
2
w0 L2 w0 x 2 w0 x 3
→ Mz = −
+
−
6
2
3L
(e) Integrate the moment for L/2 < x < L.
w0 L2 w0 x 2
w0 x 3
Mz
v2 ' ' =
=−
+
−
EI zz
6 EI zz 2 EI zz 3LEI zz
w0 L2
w0 x 3
w0 x 4
v2 ' = −
x+
−
+ C3
6 EI zz
6 EI zz 12 EI zz
w0 L2 2 w0 x 4
w0 x 5
v2 = −
x +
−
+ C3 x + C4
12 EI zz
24 EI zz 60 EI zz
18
CE 206 - Deflection of Beams
10/17/01
(f) To determine C 3 and C 4 , the continuity
conditions at L/2 are applied...
v1 ( L / 2) = v 2 ( L / 2)
v1 ' ( L / 2) = v 2 ' ( L / 2)
where
w0 L4
v1 ( L / 2) = −
48EI zz
3w0 L4
L
v 2 ( L / 2) = −
+ C1 + C 2
160 EI zz
2
7 w0 L3
v1 ' ( L / 2) = −
96 EI zz
13w0 L3
v 2 ( L / 2) = −
+ C1
192 EI zz
Setting v1 ' ( L / 2) = v 2 ' ( L / 2) :
7 w0 L3
13w0 L3
w0 L3
−
=−
+ C1 → C1 = −
96 EI zz
192 EI zz
192 EI zz
Setting v1 ( L / 2) = v 2 ( L / 2) : (and substituting C1 from
above)
w0 L4
41w0 L4
w0 L4
−
=−
+ C2 → C2 =
48EI zz
1920 EI zz
1920 EI zz
Therefore, for x > L/2:
w0 L4
w0 L2 x 2
w0 x 4
w0 x 5
w0 L3
+
−
−
x+
v2 ( x) = −
12 EI zz 24 EI zz 60 LEI zz 192 EI
1920 EI zz
19
CE 206 - Deflection of Beams
10/17/01
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 E , I zz are constant
To the left of the concentrated load,
( EI zz v1 ' ' )' ' = 0 → v iv = 0
v1 ' ' ' = C1
v1 ' ' = C1 x + C 2
C1 2
x + C2 x + C3
2
C1 3 C 2 2
v1 =
x +
x + C3 x + C4
6
2
B.C.: At x=0,
v1 ' =
v1 (0) = 0 = C 4
→
C4 = 0
v1 ' ' (0) = 0 = C 2
→
C2 = 0
so,
v1 =
C1 3
x + C3 x
6
20
CE 206 - Deflection of Beams
10/17/01
To the right of the concentrated load:
( EI zz v 2 ' ' )' ' = 0 → v 2 = 0
iv
v 2 ' ' ' = C5
v 2 ' ' = C5 x + C6
C5 2
x + C6 x + C7
2
C
C
v2 = 5 x 3 + 6 x 2
6
2
+ C 7 x + C8
v2 ' =
B.C. At x=L,
L3
L2
+ C 7 L + C8
v 2 ( L) = 0 = C 5 + C 6
6
2
v 2 ' ' ( L) = 0 = C 5 L + C 6
→
C 6 = −C 5 L,
L2
C8 = C5 − C 7 L
2
so
x 2 L3 
 x3
v 2 = C 5  − L +  + C 7 ( x − L)
2
3
 6
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.
21
CE 206 - Deflection of Beams
10/17/01
At the load point,
(i) Continuity of displacements:
L3
L
11L3
L
 L
 L
v1   = v 2   → C1 + C 3 =
C5 − C7
48
2
48
2
2
2
(ii) Continuity of slopes:
L2
3L2
 L
 L
v1 '   = v 2 '   → C1 + C 3 = −
C5 + C7
8
8
2
2
(iii) Continuity of moments:
M 1 ( L / 2 ) = M 2 ( L / 2)
→ EI zz v1 ' ' ( L / 2) = EIv 2 ' ' ( L / 2)
→ v1 ' ' ( L / 2 ) = v 2 ' ' ( L / 2 )
since E , I zz are constant. Therefore,
v1 ' ' ( L / 2) = v 2 ' ' ( L / 2) → C1
22
L
L
= −C 5 → C1 = −C 5
2
2
CE 206 - Deflection of Beams
10/17/01
(iii) Shear equilibrium:
From equilibrium, at a point load, -P,
∆V y = V y2 ( L / 2) − V y1 ( L / 2) = − P
But,
V y1 = ( EI zz v1 ' ' )' = EI zz C1
V y2 = ( EI zz v 2 ' ' )' = EI zz C5
so
EI zz C5 − EI zz C1 = − P
Using these results together, obtain
C1 =
P
P
, C5 = −
2 EI zz
2 EI zz
Substituting C1 and C 5 into the displacement and
slope continuity conditions,
L3  P  L
11L3  − P  L
 + C 3 =
 − C 7


48  2 EI zz  2
48  2 EI zz  2
L2  P 
3L2  − P 
 + C 3 = −
 + C 7


8  2 EI zz 
8  2 EI zz 
PL2
3PL2
, D3 = −
→ C3 = −
16
16 EI zz
23
CE 206 - Deflection of Beams
10/17/01
Substituting all of these constants into the solutions,
P
PL2
3
v1 =
x −
x
12 EI zz
16 EI zz
P  x3
x 2 L3  3PL2
−L + −
v2 = −
( x − L)

2 EI zz  6
2
3  16 EI zz
The displacement at any point of the beam is given
by
v1 ( x )
v( x ) = 
v 2 ( x )
0 ≤ x ≤ L/2
L/2 < x ≤ L
Maximum displacement occurs at L/2, where
PL3
v ( L / 2) = −
48 EI zz
The - sign means that the displacement is downward,
consistent with the downward loading.
24
CE 206 - Deflection of Beams
10/17/01
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.
25
CE 206 - Deflection of Beams
10/17/01
S.P. 7: Analyze the statically indeterminate beam
shown.
(a) Conduct the static analysis separately, and
incorporate the moments as unknowns:
∑F
∑M
y
= A y + B y − w0 L = 0
( B)
Z
= − Ay L − Az + B z + wo L2 / 2
wo L
1
→ Ay = − ( Az − B z ) +
L
2
w L 1
B = o + (A − B )
y
z
L z
2
Note that Ay and B y have two components:
(i) Effect of load if Az , B z are zero.
(ii) Effect of Az , B z .
26
CE 206 - Deflection of Beams
10/17/01
Next, cut a FBD at x and obtain the moment.
∑M
( x)
Z
wo x 2
 wo L 1

= M z ( x ) − Az − 
− ( Az − B z ) x +
2
L
 2

=0
wo Lx wo x 2
x
x

→ M z ( x) =
−
+ Az 1 −  + B z
2
2
L
 L
Therefore, the curvature is given by
d 2 v M z ( x)
=
2
EI zz
dx
1
=
EI zz
 w0 Lx wo x 2
x

−
+
−
1
A

 + Bz
z
 2
2
 L

x
L 
Integrate twice. Before doing this, note that the
boundary conditions are
v ( 0) = v ' ( 0) = v ( L ) = v ' ( L ) = 0
Here, let's use indirect integration.
27
CE 206 - Deflection of Beams
10/17/01
The first integration yields

dv
1  w0 Lx 2 wo x 3
x2 
x
=
−
+ Az  x −  + B z  + C1

dx EI zz  4
6
2L 
L

But, v'(0)=0 implies that C1 = 0, so this reduces to
dv

x2 
1  w0 Lx 2 wo x 3
=
−
+ Az  x −  + B z

dx EI zz  4
6
2L 

x
L 
The second integration yields
 x2 x3 
1  w0 Lx 3 wo x 4
x2 
−
+ Az  −  + B z
+ C2
v( x ) =


EI zz  12
24
2L 
 2 6L 
But, v(0)=0 implies that C 2 = 0 . So, the displacement
is
 x2 x3 
x2 
1  w0 Lx 3 wo x 4
v( x ) =
−
+ Az  −  + B z

EI zz  12
24
2 L 
 2 6L 
At this point the boundary conditions at x=0 have
been satisfied.
28
CE 206 - Deflection of Beams
10/17/01
We still don't know Az and B z , but we still have the
conditions v ( L) = 0, v ' ( L) = 0. Hence,
wo L4
Az L2 B z L2
+
+
=0
v ( L) =
24 EI zz 3EI zz 6 EI zz
wo L3
AL
B L
+ z + z =0
v ' ( L) =
12 EI zz 2 EI zz 2 EI zz
Solving these two equations simultaneously yields
w0 L2
Az = B z = −
12
Therefore, the moments are
wo Lx wo x 2 wo L2
−
−
M z ( x) =
2
2
12
and the displacements are
w0 Lx 3
wo x 4
wo L2 x 2
−
−
v( x ) =
12 EI zz 24 EI zz 24 EI zz
These functions are plotted below.
29
CE 206 - Deflection of Beams
10/17/01
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).
30
CE 206 - Deflection of Beams
10/17/01
S.P. 8: Solve the problem of S.P. 7 by integrating the
4th order differential equation
wo
EI zz v = − wo → v = −
EI zz
iv
iv
First Integration:
wo x
+ C1
v' ' ' = −
EI zz
Second Integration:
wo x 2
+ C1 x + C 2
v' ' = −
2 EI zz
Third Integration:
wo x 3
x2
+ C1
+ C2 x + C3
v' = −
6 EI zz
2
Fourth Integration:
wo x 4
x3
x2
+ C1
+ C2
+ C3 x + C 4
v=−
24 EI zz
6
2
4 constants of integration must be determined from
the boundary conditions.
31
CE 206 - Deflection of Beams
10/17/01
BC at x=0:
v ( 0) = 0 = C 4
→
C4 = 0
v ' ( 0) = 0 = C 3
→
C3 = 0
So, the solution reduces (at this point to
wo x 4
x3
x2
+ C1 + C 2
v=−
24 EI zz
6
2
BC at x=L:
wo L4
L3
L2
v ( L) = 0 = −
+ C1 + C 2
24 EI zz
6
2
wo L3
L2
v ' ( L) = 0 = −
+ C1 + C 2 L
6 EI zz
2
Solving these two equations for C1 and C 2 ,
w L
C1 = o ,
2 EI zz
wo L2
C2 = −
12 EI zz
The solution then becomes
w0 Lx 3
wo x 4
wo L2 x 2
−
−
v( x ) =
12 EI zz 24 EI zz 24 EI zz
in agreement with the solution of S.P. 7.
32
CE 206 - Deflection of Beams
10/17/01
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,
( EI zz v1 ' ' )' ' = 0 → v iv = 0
v1 ' ' ' = C1
v1 ' ' = C1 x + C 2
C1 2
x + C2 x + C3
2
C
C
v1 = 1 x 3 + 2 x 2 + C 3 x + C 4
6
2
v1 ' =
33
CE 206 - Deflection of Beams
10/17/01
B.C.: At x=0,
v1 (0) = 0 = C 4
→
C4 = 0
v1 ' ' (0) = 0 = C 2
→
C2 = 0
so,
C1 3
v1 =
x + C3 x
6
To the right of the concentrated load:
( EI zz v 2 ' ' )' ' = 0 → v 2 = 0
iv
v 2 ' ' ' = C5
v 2 ' ' = C5 x + C6
C5 2
x + C6 x + C7
2
C
C
v 2 = 5 x 3 + 6 x 2 + C 7 x + C8
6
2
v2 ' =
B.C. At x=L,
L3
L2
v 2 ( L) = 0 = C 5 + C 6 + C 7 L + C8
6
2
L2
v 2 ' ( L) = 0 = C5 + C 6 L + C 7
2
(These won't be quite as easy to solve as in S.P. 6, so
we'll try a more formal approach.)
34
CE 206 - Deflection of Beams
10/17/01
The continuity conditions at the load point are the
same as in S.P. 6. Namely,
(i) Continuity of displacements:
v1 ( L / 2) = v 2 ( L / 2)
L3
L
L3
L2
L
→ C1 + C 3 = C5
+ C6
+ C 7 + C8
48
2
48
8
2
(ii) Continuity of rotations:
v1 ' ( L / 2) = v 2 ' ( L / 2)
L2
L2
L
→ C1 + C 3 = C 5 + C 6 + C 7
8
8
2
(iii) Continuity of moments:
v1 ' ' ( L / 2) = v 2 ' ' ( L / 2)
L
L
= C5 + C6
2
2
(iv) Shear equilibrium:
→ C1
EI zz ( v1 ' ' '−v 2 ' ' ' ) = P → C1 − C 5 = P / EI zz
To solve, let's organize these algebraic boundary and
continuity condition equations into a matrix format..
35
CE 206 - Deflection of Beams
10/17/01


L3
L2
L
0
0
1


6
2

 C1   0 
2
L
0 0
L
1
0    0 
2


 C 2  
3
3
2
L
L −L
L L
 C   0 
− 1 5 =
 48 2 − 48 − 8
   0 
2
 L2

2
C 6  

−
L
L
0


−
− 1 0 C 7  
1

8
2
8
   P 
L
L
 C8   EI 
−
−
0
1
0
0
zz
2

2
 1 0 − 1
0
0
0 
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,

C1  5 / 16
C  − L2 / 32 

 2 
C 5  − 11 / 16  P
 =

/
2
L
C
 6 
 EI zz
C 7  − 5L2 / 32 
   3

C8   L / 48 
36
CE 206 - Deflection of Beams
10/17/01
Then, the solution is
 5 x 3 L2  P
− x
v1 = 
 96 32  EI zz
0 ≤ x ≤ L/2
 11x 3 Lx 2 5L2 x L3  P
+
−
+ 
v 2 = −
4
32
48  EI zz
 96
37
L
<x≤L
2
CE 206 - Deflection of Beams
10/17/01
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!
38