Chapter 8 Displacement of Beams
It is often the case that the amount a beam deflects or rotates due to the applied transverse
loads has a limit placed on it. For this reason we now need to determine the deflection and
slope of a transversely loaded beam. This can be done in one of three ways: (1) Double
integration method; (2) Moment area method and (3) Energy method.
8.1 DOUBLE INTEGRATION METHOD (SI&4th: 569-599; 5th: 569-599)
Elastic Curve Equation
Consider a loaded beam as shown in Fig. 8.1:
y, v
w
P
M
x
θ
dx
x
Fig. 8.1 Loaded beam indicating its deflection and slope due to the applied loads
Look at a FBD of element dx, and consider only the bending moment that it is experiencing:
O
dθ
R
R
M
N. A
y
dθ
M
.
dx
Fig. 8.2 The curvature of deflection
θ
dv
dx
Fig. 8.3 Infinitesimal segment dx showing angle
and vertical displacement
From Engineer's Theory of Bending (ETB), Eq. (6.9), we know that a beam under an applied
bending moment deflects with a curvature equal to the radius of a circle (arc), and that this
radius is related to the applied bending moment by:
1 M
=
(8.1)
R EI
From Fig. 8.2, we can approximately compute arc dx = Rdθ , therefore,
1 dθ
(8.2)
=
R dx
where θ can be considered to be the slope of the beam. But from the a more detailed diagram showing
Fig. 8.3, when dx approximates to zero and the slope is small enough, we have relationship as
dv
= tan θ ≈ θ
(8.3)
dx
Lecture Notes of Mechanics of Solids, Chapter 8
1
Combining Eqs. (8.2) and (8.3), we can obtain:
1 dθ d 2 v
=
=
(8.4)
R dx dx 2
and then substituting the part from ETB, Eq. (8.1), it gives Elastic Curve Equation:
d 2v M
=
(8.5)
dx 2 EI
d 2v
= M (x )
(8.6)
dx 2
here EI is referred to as flexural rigidity. Furthermore, from the relationship between distributed
load w(x) and shear force V(x) established in Chapter 5 (Eqs. (5.2) and (5.2)), one can have
d 3v
EI 3 = V ( x )
(8.7)
dx
or
EI
and
EI
d 4v
= − w( x )
dx 4
(8.8)
Knowing the material and cross sectional properties of the beam, i.e. flexural rigidity(EI),
Eq. (8.6) can be integrated ONCE to give an equation for the slope (θ) and TWICE to give
an equation for the displacement (v) of the beam as a function of x as:
dv
M (x )
dv
θ=
=∫
dx + C or EIθ = EI
= M ( x )dx + C
(8.9)
dx
EI
dx ∫
M (x )
v = ∫∫
dxdx + Cx + D or EIv = ∫∫ M (x )dxdx + Cx + D
(8.10)
EI
Integration of Macaulay’s Function
Recalling Chapter 5, we adopted Macaulay’s function, Eq. (5.11), as
for x < a
 0
(n ≥ 0)
(8.11)
=
n
(x − a ) for x ≥ a
in the expression of the shear force and bending moment equations. In order to compute the
deflection v and slope θ = dv/dx from Eq. (8.5) or (8.6), we need to integrate Macaulay’s function as
x−a
∫
n
n
x − a dx =
x−a
n +1
+ C'
(8.13)
n +1
where C’ is constant of integration and can be determined from the kinematic boundary conditions.
Kinematic Boundary Conditions
The boundary conditions of the Cantilever and Simply Supported Beams can be seen from Fig. 8.4.
At x=0, v=0 and θ=dv/dx=0
Cantilever
Beam
x
At x=0, v=0
At x=L, v=0
Simply
Supported
Beam
x
L
Fig. 8.4 Kinematic boundary conditions
Lecture Notes of Mechanics of Solids, Chapter 8
2
Example 8.1 Determine the slope and displacement equations of the Simply Supported (SS)
beam with a point load P.
F.B.D. (global equilibrium)
P
a
I
A
θA
vmax
θB
F.B.D. (Section I-I)
P
L/2
B
I
A
x
L
RAY=(1-a/L)P
I
o
I
x
RBY=Pa/L
M(x)
V(x)
RAY=P/2
Step 1: Determine the ground reactions at supports A and B;
RAY = P/2 and RBY = P/2
Step 2: Bending moment equation via equilibrium for FBD of Section I-I.
By cutting the beam just before the RHS (Section I-I), the bending moment can be determined
as discussed in Chapter 5. Take moments about RHS:
+
P 1
L
∑MO = 0 = − 2 x + P x − 2
1
P 1
L
∴ M (x ) =
x −P x−
2
2
+ M (x ) = 0
1
Step 3: Double Integration for the elastic curve equation.
Substituting M(x) into the elastic curve equation Eq. (8.6), gives that:
d 2v P 1
L
d 2v
x −P x−
EI 2 = M ( x ) → EI 2 =
2
2
dx
dx
Integrate once:
dv P
EIθ = EI
=
x
dx 4
2
P
L
−
x−
2
2
1
2
+ C (slope equation)
3
P
P
L
3
x −
x−
+ Cx + D
(elastic curve equation)
Integrate again: EIv =
12
6
2
where C and D are the constants of the integration. To determine them one needs to use Kinematic
Boundary Conditions, which are from the known displacements and rotations of the beam.
Step 4: Determine the integration constants based on Kinematic Boundary Conditions
Our beam is simply supported at both ends, so kinematic boundary conditions are, as Fig. 8.4,
• when x = 0, v = 0, (recall the definition of Macaulay’s function as in Eq. (8.11))
P
P
EI × 0 = × 0 3 − × 0 + C × 0 + D = 0 ∴D = 0
12
6
• when x = L, v = 0
3
3
P
P
L
P
P
L
PL2
3
3
EI × 0 =
L −
L−
+ CL + 0 = (L ) −  L −  + CL = 0 ∴ C = −
12
6
2
12
6
2
16
Step 5: Express the Slope and Elastic Curve Equations respectively:
dv P
=
x
Slope Equation: EIθ = EI
dx 4
P
x
Elastic Equation: EIv =
12
3
2
P
L
−
x−
2
2
P
L
−
x−
6
2
3
−
2
−
PL2
16
PL2
x
16
PL3
(downwards)
48 EI
PL
(counter-clockwise)
θB =
16 EI
At x = L/2, the deflection reaches the maximum as v max = −
At x = 0,
θA = −
PL
(clockwise) and at x = L,
16 EI
Lecture Notes of Mechanics of Solids, Chapter 8
3
8.2 MOMENT-AREA METHOD (SI&4th: 600-613, 5th: 600-613)
The Moment-Area Method uses the Elastic Curve Equation derived above, but the integration
is done graphically and by doing so the kinematic boundary conditions are not considered.
Let’s look at elastic curve equation (8.5):
d 2v M
=
dx 2 EI
To integrate this equation graphically, you firstly need to draw the Bending Moment M(x)
diagram and then the M(x)/EI diagram. So look at a beam with arbitrary loadings as illustrated
in Fig. 8.5:
F1
F2
w(x)
Loading
Diagram
x1
x2
M(x)
Bending
Moment
Diagram
x1
x
x2
M(x)/EI
A=θ2 - θ1
M(x)/EI
Diagram
x1
x2
x
Fig. 8.5 Bending moment and M/EI diagrams for beam with arbitrary loading
1st Theorem of Moment Area
Integrating the elastic curve equation with respect to x, between two points x1 and x2, gives:
2
x2 d v
x2 M
∫ x1 dx 2 dx = ∫ x1 EI dx
which can be reduced to:
x
x2 M
 dv  2
 dx  = ∫ x1 EI dx
x1
(8.14)
x2 M
 dv 
 dv 
dx
  −   = θ 2 − θ1 = ∫ x
1 EI
 dx  x2  dx  x1
(8.15)
This is
This equation gives the change in slope of the beam between x1 and x2. It is represented by the
area in the M/EI diagram between x1 and x2, and this equation is called the 1st THEOREM OF
MOMENT AREA. Note that the AREA should be considered in an algebraic sense, i.e. can
be positive or negative.
In this theorem, if dv/dx is known at x1, dv/dx at x2 can be very easily found via Eq. (8.15).
A useful side effect of this is that if I varies along the length of the beam, it can easily be
accommodated for, as to be shown in Example 8.2 below.
Lecture Notes of Mechanics of Solids, Chapter 8
4
Example 8.2 Look at a cantilever beam where I = I0 for the left half of beam and I = I 0 / 2,
for the right half. Find the slope θ = dv/dx at the end C.
L/2
Loading
Diagram
A
I0
M(x)
B
L/2
P
I0/2
C
L/2
Bending
Moment
Diagram
L
x
-PL/2
-PL
M(x)/EI
M(x)/EI
Diagram
-PL
2EI0
-PL
EI0
L/2
1
2
L
x
3
-PL
EI0
We know that at x = 0, dv/dx = 0, so use the 1st Theorem of Moment Area Eq. (8.15) gives that:
M (x )
 dv 
 dv 
diagram between x=0 and x=L
  −   = θ L − θ 0 = Area under
EI
 dx  L  dx  0
which can be computed by adding those three sub areas as shown,
 PL
L 1 L PL
1 L PL 
5 PL2
 dv 
so:   − 0 = A1 + A2 + A3 = −
× − × ×
− × ×
=
−

8 EI 0
 dx  L
 2 EI 0 2 2 2 2 EI 0 2 2 EI 0 
which gives that :
5 PL2
 dv 
θC =   = −
8 EI 0
 dx  L
2nd Theorem of Moment Area
Equation (8.15) only gives the change of slope between any two points, to determine the
displacement at some points along the beam, the second moment theorem must be applied. To
find displacement let’s return to elastic curve equation (8.5):
d 2v M
=
dx 2 EI
Multiply both sides by x and integrate between x1 and x2,
2
x2
x2 M
x2 d v
x2 M
 dv 
∫ x1 dx 2 xdx = ∫ x1 EI xdx i.e. ∫ x1 xd  dx  = ∫ x1 x EI dx
Integrating this by parts:
(8.16)
∫ pdq = pq − ∫ qdp
where p and q are functions, gives:
x2
∫x
1
x
x2 dv
x2 M
 dv   dv  2
xd   =  x  − ∫
dx = ∫ x dx
x1 dx
x1
EI
 dx   dx  x1
(8.17)
x
x2 M
dv
 dv
 2
x
v
x dx
−
=
=
dx
,
Eq.
(8.17)
becomes:
∫
∫ x1 dx
 dx

1
x1
EI
x1
As a result, we have
But, [v ]xx
2
x2
Lecture Notes of Mechanics of Solids, Chapter 8
5
 dv 

x2 M
 dv 
  x 2 −   x1  − [v2 − v1 ] = ∫ x x dx
1
EI
 dx  x1 
 dx  x2
(8.18)
[θ 2 x2 − θ1 x1 ] − [v2 − v1 ] = Ax
M(x)/EI
x
A
M(x)/EI
Diagram
x1
x
x2
Fig. 8.6 The 2nd theorem of moment area for finding deflection
as the 2nd THEOREM OF MOMENT AREA. It can give the change in deflection (v1-v2)
between any two points x1 and x2 in terms of the change in slope and the first moment of the
area in the M/EI diagram as shown in Fig. 8.6.
Example 8.3 To see how this works look at the above example, but this time we require to
determine the displacement v = ? at the tip.
Kinematic conditions:
at x = 0 , v = 0 , dv/dx = 0 and
at x = L , dv/dx = -5PL2/8EI0,
Setting x1=0 and x2=L in this example, substituting these values into Eq. (8.18) gives:
M(x)/EI
L/2+L/6=2L/3
L/4
xM(x)/EI
Diagram
L/2
1
-PL
2EI0
-PL
EI0
2
L
x
3
-PL
EI0
L/6
 5 PL2

 PL L  L  1 L PL  L  1 PL L  2 L
 × −  ×
 ×
× L − 0 − [v − 0] = −
×  × −  ×
−
EI
8
2
2
4
2
2
2
6
2
2
EI
EI
EI
0
0
0 
0




 3


When this is solved it gives that:
3 PL3
v=−
8 EI 0
8.3 ENERGY METHOD (SI&4th: 712-717, 768-774; 5th: 712-717, 768-774)
Consider a generalized beam loaded as shown in Fig. 8.7.
P1
P2
w(x)
MC
A
B
D
C
MB
θC
x
dx
Fig. 8.7 Beam under arbitrary loads
Lecture Notes of Mechanics of Solids, Chapter 8
6
Because of the loading condition this beam has a bending moment distribution along its length. In
Section 3.4, an equation, Eq. (3.8), for the stored strain energy of a structure was defined as:
σ2
(8.19)
U =∫
dV
V 2E
In the transversely loaded beam, the normal stress distribution is given by Engineer’s Theory
of Bending, Eq. (6.10), such that:
M (x )
σx = −
y
(8.20)
I
Substituting Eq. (8.20) into Eq. (8.19) gives:
M 2 (x ) y 2
U =∫
dV
(8.21)
V
2 EI 2
but for a transversely loaded beam, dV = dA×dx, thus:
1
M 2 (x ) y 2
(8.22)
U= ∫ ∫
dAdx
2 L A
EI 2
because M(x), E and I are constant for a specific cross section then:
1
M2
U= ∫
y 2 dAdx
(8.23)
2 L EI 2 ∫ A
Since I = ∫ y 2 dA , then the total Strain Energy Stored in a Straight Beam is given by:
A
M2
(8.24)
dx
L 2 EI
And for a Circular Beam, the equation becomes:
M 2 (θ)
(8.25)
U =∫
Rdθ
θ 2 EI
To determine the displacement at the point of application of the load, Castigliano's 2nd
Theorem is used. So differentiating the total Strain Energy with respect to the applied load P
gives the desired deflection as:
L M ( x ) ∂M ( x )
∂U
(8.26)
=∫
vP =
dx
0
∂P
∂P
EI
In order to determine the slope of tangent θ at a point on elastic curve, the partial derivative of
the internal bending moment M(x) with respect to an external bending moment M’ acting at
the point must be found, as
L M ( x ) ∂M ( x )
∂U
(8.27)
θM ′ =
=∫
dx
0
∂M ′
EI ∂M ′
For example, at point C in Fig. 8.7, one can find the slope at C by formulating as
L M ( x ) ∂M ( x )
∂U
θC =
=∫
dx
(8.28)
0
∂M C
EI ∂M C
U =∫
Example 8.4 Determine vertical deflection for a simply supported beam with a central load P.
Step 1: Bending moment equation From Example 8.1, the bending moment equation is:
M (x ) =
P 1
L
x −P x−
2
2
1
Lecture Notes of Mechanics of Solids, Chapter 8
7
F.B.D. (global equilibrium)
P
L/2
C
A
I
B
I
L
RAY=P/2
RBY=P/2
Step 2: Compute the total strain energy U
Because of Macaulay’s notation we have to do the following:
U =∫
L
M 2 (x )
dx =
2 EI
L/2
∫
0
M 2 (x )
M 2 (x )
dx + ∫
dx
2 EI
2
EI
L/2
L
and substituting for the bending moments it gives:
2
2
 Px 
P

(L − x )

L/2 
L 
2 
2


 dx
U= ∫
dx + ∫
2 EI
2 EI
0
L/2
doing this integration gives the total bending strain energy as:
P 2 L3
P 2 L3
P 2 L3
U=
+
=
192 EI 192 EI 96 EI
Step 3: Castigliano's 2nd Theorem
The displacement is then found by Castigliano's Theorem:
∂U
∂  P 2 L3  PL3
=

vP =
=
∂P ∂P  96 EI  48EI
which is the displacement at the point of application of P in the direction of P.
Remarks
We may observe that the deflection vj of a beam at a given point C can be obtained by direct
application of Castigliano’s theorem only if a real load Pj happens to be applied at C in the
direction in which vj is to be determined. When no real load is applied at Cj, or when a real load
P is applied in a direction other than the desired one, we need to apply a fictitious or virtual
load Qj at Cj along the direction in which the deflection vj is to be determined and use
Castigliano’s theorem to obtain deflection vj, similarly to the approach stated in Chapter 3, as
∂U (P ,Q j )
(8.29)
vj =
∂Q j
Keep in mind that the internal strain energy here contains the contributions from both actual
load P and virtual load Qj. After computing the partial derivative with respect to Qj, then
make Qj = 0 in Eq. (8.29).
The slope θj of a beam at point Cj may be determined in a similar manner by applying a fictitious
couple Mj at Cj, then computing the partial derivative as
∂U P , M j
θj =
(8.30)
∂M j
(
)
and making Mj = 0 in the expression obtained.
Lecture Notes of Mechanics of Solids, Chapter 8
8
Example 8.5 The cantilever beam AB supports a uniformly distributed load w as shown.
Determine the deflection vB and slope θB at the free end B.
F.B.D. (global equilibrium)
w
I
w
MA
B
A
A
L
I
L
RAY
Virtual load to
find deflection vB
B
x
QB
Part I: Determine Deflection vB
Since there is no real concentrated load at B along vertical direction, a fictitious load QB must
be applied at the point along the desired direction as shown.
Step 0: Ground reactions due to both w and QB:
∑ M A = 0 = M A − QB L − (wL )(L / 2) = 0
+
+ ↑ ∑ Fy = 0 = R AY − QB − wL = 0
∴ M A = wL2 / 2 + QB L
∴ R AY = wL + QB
Step 1: Bending moment equation (via Section I-I)
w 2
0
1
x + M (x ) = 0
+ ∑ M O = 0 = M A x − R AY x +
2
M (x ) = − M A x
0
1
+ R AY x − w x
2
(
)
/ 2 = − wL2 / 2 + Q B L x
0
+ (wL + QB ) x − w x
1
2
/2
Step 2: Compute the total strain energy U in terms of both real (w) and virtual load (QB)
1
M 2 (x )
w 2
0
1

dx =
x  dx
− wL2 / 2 + Q B L x + (wL + Q B ) x −
∫

L 2 EI
2 EI L 
2

Step 3: Using Castigliano’s Theorem
∂M ( x , w,Q B )
∂U (w,Q B )
M 2 ( x , w,Q B )
∂
1
dx
=
dx =
M
vB =
∫
∫
L
L
∂Q B
∂QB
∂Q B
EI
2 EI
∂M ( x , w,QB )
0
1
But
= −L x + x
∂QB
Substituting for M and ∂M ∂QB into the previous equation and setting QB = 0, we have
(
U (w,Q B ) = ∫
vB =
1
EI
1
=
EI
∫L
M
1
∂M
dx =
EI
∂QB
  L2

∫ L  −  w 2 + 0  x

2
)
0
+ (wL + 0) x −
1
(
w 2 
x
−L x

2

0
+ x
1
)dx
L
 wL3 3wL2

1
3wL 2 w 3 
L2
w 2



(
)
w
wLx
x
x
L
dx
x
x − x dx
−
=
−
+
−
+
−
∫ L  2
2 
2
2
2 
EI ∫0  2
wL4
vB =
(“+” means the same direction as QB)
8EI
Part II: Determine slope θB
F.B.D. (global equilibrium)
w
I B
MA
x
A
RAY
L
I MB
Virtual moment to
find slope θB
Slope corresponds to a couple in Castigliano’s 2nd theorem. It is hence necessary to apply a
virtual moment MB at the B as shown.
Lecture Notes of Mechanics of Solids, Chapter 8
9
Step 0: Ground reactions due to both w and MB:
+ ∑ M A = 0 = M A − M B − (wL ) L  = 0
2
+ ↑ ∑ Fy = 0 = R AY − wL = 0
∴ M A = wL2 / 2 + M B
∴ R AY = wL
Step 1: Bending moment equation (via Section I-I)
w 2
0
1
x + M (x ) = 0
+ ∑ M O = 0 = M A x − R AY x +
2
w 2
w 2
0
1
0
1
M ( x ) = − M A x + R AY x −
x = − wL2 / 2 + M B x + wL x −
x
2
2
Step 2: Total strain energy U in terms of both real force (w) and virtual moment (MB)
(
)
1
M 2 (x )
w 2
0
1

2
2
dx =
wL
/
M
x
wL
x
x  dx
−
+
+
−
B
∫
L 2 EI
2 EI L 
2

Step 3: Castigliano’s Theorem
∂M ( x , w, M B )
∂U (w, M B )
M 2 ( x , w, M B )
∂
1
dx
=
dx =
M
θB =
∫
∫
L
L
∂M B
∂M B
∂M B
EI
2 EI
∂M ( x , w, M B )
0
but
=− x
∂M B
Substituting for M and ∂M ∂M B into the previous equation and then setting MB = 0, we have
(
U (w, M B ) = ∫
1
θB =
EI
=
1
EI
∫L
∫L
∂M
1
M
dx =
∂QB
EI
  L2

 −  w + 0 x

  2

 
0
  L2

− w + M B  x

  2

 
∫L
1
+ wL x −
θB =
2
)
0
1
+ wL x −
1
w 2 
x (− 1)dx =

2
EI

L
∫
0
(
w 2 
x
− x

2

0
)dx
 wL2
w 2
wL3
dx =

wLx
x
−
+
 2
2 
6 EI

3
wL
(the same rotational direction as MB)
6 EI
So far we have just been looking at beams that are statically determinate, we now need to
look at the cases when the beams are statically indeterminate, that is there are more unknown
reaction forces than equations of statics. In these cases we need to come up with as many
compatibility equations as are necessary to solve the problem.
8.4 STATICALLY INDETERMINATE BEAMS (SI&4th: 614-647; 5th: 614-647)
A beam is statically indeterminate when there are more unknown support loads (forces &
moments) than equations of statics. Because these excess support loads have associated with
them excess boundary conditions, we can use these to solve the problem. For example, for a
propped cantilever as shown in Fig. 8.8:
F.B.D. (global equilibrium)
P
2L/3
MA
RAY
I
B
A
I
L
RBY
Fig. 8.8 Statically indeterminate cantilever beam with load P
Lecture Notes of Mechanics of Solids, Chapter 8
10
Three unknown reactions, RA, RB and MA ;
Two equations of statics (ΣFy=0 and ΣM=0)
Three kinematic boundary conditions, at x = 0, v =0 and dv/dx = 0, at x = L, v = 0.
8.4.1 INTEGRATION METHOD TO SOLVE STATICALLY INDETERMINATE BEAMS
When employ the integration method as given in Section, 8.1, two of these kinematic boundary
conditions are necessary for determining the constants of integration; the third is used with the
two equations of statics to solve for the three reaction loads.
Example 8.6 Determine the reaction loads in the indeterminate beam with a point force P
applied at 2/3L, as shown in Fig. 8.8.
Step 1: Global equilibrium from statics:
+ ↑ ∑ Fy = 0 = R AY + R BY − P = 0
+
(8.31)
∑ M A = 0 = M A − 2 PL / 3 + RBY L = 0
(8.32)
Step 2: Bending moment equation
Take moments about Section I-I by cutting just before RHS, as shown in Fig. 8.8:
+
∑MO = 0 = M A
x
0
− R AY x + P x − 2 L / 3 + M ( x ) = 0
1
1
The Moment equation is given by:
M (x ) = − M A x
0
1
+ R AY x − P x − 2 L / 3
1
Step 3: Determine Elastic Curve equation
We now need to derive the displacement equation and apply the kinematic boundary conditions:
Using the Elastic curve equation gives that:
d 2v
0
1
1
EI 2 = M ( x ) = − M A x + R AY x − P x − 2 L / 3
dx
R
dv
P
1
2
2
Integrate once:
EIθ = EI
= ∫ M ( x )dx = − M A x + AY x −
x − 2L / 3 + C
dx
2
2
MA
R AY
P
2
3
3
Integrating again: EIv = ∫∫ M ( x )dxdx = −
x +
x −
x − 2 L / 3 + Cx + D
2
6
6
Step 4: Determine the integration constants based on Kinematic Boundary Conditions
Since at x = 0 , dv/dx = 0 , then C = 0 Since at x = 0 , v = 0 ,
then D = 0
MA
R
P
2
3
3
Elastic Curve Equation: EIv = −
x + AY x −
x − 2L / 3
2
6
6
Step 5: Give an additional equation using other Kinematic Boundary Condition
Also, at x = L, v = 0, we have equation: −
MA
L
2
2
+
R AY
L
6
3
−
P
2
L− L
6
3
3
=0
(8.33)
which gives : R AY L − 3M A − PL / 27 = 0
Solving Eqs. (8.31), (8.32) and (8.33) simultaneously gives that:
RA = 13P/27,
RB = 14P/27,
MA = 4PL/27
You can then substitute these values into the above equations to obtain the slope and
displacement of the beam.
Lecture Notes of Mechanics of Solids, Chapter 8
11
8.4.2 SUPERPOSITION METHOD TO SOLVE STATICALLY INDETERMINATE BEAMS
An alternative way of solving the above problem is by using the superposition method. Using
the superposition method we can very easily generate the extra equations necessary to solve
the statically indeterminate beam. This can be done by determining the deflection on the beam
due to each of the applied loads and then add all of these displacements together.
Example 8.7 The same as Example 8.6 but using Superposition Method.
The loads and displacements in this beam are equivalent to treating it as two separate statically
determinate beams, then combining the separate displacements in the following way.
P
2L/3
B
AM
A
RAY
L
=
MA1
RBY
P
2L/3
v1
A
RAY1
B
B
+
A
MA2
L
RAY2
RBY
L
To determine the displacements v1 and v2 you can refer to standard solutions given in the
Textbook in Appendix C, 4th:800-801; 5th 800-801 or in other references.
For this example, the equation for v1 is:
2
P(2 L 3) 
2L 
v1 = −
 3L −

6 EI 
3 
The equation for v2 is :
R L3
v 2 = BY
3EI
But because at B, v = 0, then kinematic compatibility condition is
v1 + v 2 = 0
P(2 L 3) 
2 L  RBY L3
i.e.
=0
v1 + v 2 = −
 3L −
+
6 EI 
3  3EI
and when you solve for this you get that:
14
R BY =
P
27
2
8.4.3 CASTIGLIANO’S METHOD FOR STATICALLY INDETERMINATE BEAMS
The reactions at supports of a statically indeterminate elastic structure may be determined by
Castigliano’s 2nd theorem as Eq. (8.26) or Eq. (8.27), in which the redundant reaction is treated as
an unknown load Rj. We firstly calculate the strain energy U of the structure due to the combined
action of the given loads and the redundant reaction Rj. Observing that the partial derivative
∂U ∂R j , represents the deflection (or slope) at the support. We then set this derivative equal to
zero (because of zero deflection at the support) and solve the equation obtained for the redundant
reaction. The remaining reactions may be obtained from the equation of statics.
Lecture Notes of Mechanics of Solids, Chapter 8
12
v2
Example 8.8 Determine the reactions at the supports for the prismatic beam and loading shown.
I
w
w
B
A
L
=
A
B
I
RAY
L
The beam is statically indeterminate to the first degree (i.e. one redundant reaction). We
consider the reaction at A as redundant and release the beam from the support. The reaction
RAY is now considered as an unknown load as shown and will be determined from the
condition that the deflection vA must be zero. Note that as an unknown, RAY in this case is in
effect a real load.
Step 1: Bending moment equation (via Section I-I)
w 2
1
x + M (x ) = 0
+ ∑ M O = 0 = − R AY x +
2
Step 2: Total strain energy U in terms of RAY
∴ M ( x ) = R AY x −
1
w
x
2
2
1
M 2 (x )
w 2
1

U (w, R AY ) = ∫
dx =
R AY x −
x  dx
∫

L 2 EI
L
2 EI
2


Step 3: Castigliano’s Theorem
∂U (w, R AY ) 1
∂M ( x , w, R AY )
1
∂M
vA =
M
=
dx =
M
dx
∫
∫
L
L
EI
∂R AY
EI
∂R AY
∂R AY
∂M ( x , w, R AY )
∂M
1
but
=
= x
∂R AY
∂R AY
Substituting for M and ∂M ∂R AY into the previous equation, we have
2
( )dx = EI1 ∫
1
∂M
dx =
EI
∂QB
w 2
1

∫L
∫ L  R AY x − 2 x  x
Step 4: Kinematic compatibility condition
vA = 0
vA =
1
EI
M
1
L
0
w 3

2
 R AY x − x dx
2 

L
R AY L3 wL4
1  R AY 3
w 4
x
x
∴vA =
=
−
=0
−
EI  3
2 × 4  0
3
8
∴ R AY = 3wL / 8 ↑
From the conditions of equilibrium for the beam, we find that the reaction at B consists of
following force and bending couple:
R BY = 5wL / 8 ↑
M B = wL2 / 8
Lecture Notes of Mechanics of Solids, Chapter 8
13
x
Table 8.1 Comparison of bending beams with axially loaded bars and torsional shafts
Axial Loaded Bar/Rod
Force F (N)
Load Type
Torsional Shaft
Torque T (Nm)
Tension or Compression
+T
+F
Sign
Convention
Of Internal
Load
+
Right-Hand Rule
Bending moment
+φ
+
+F
Tension
Bending Beams
Transverse Force P (N) or/and
Bending moment M (Nm)
+ve M
-ve M
+T
-F
-T
-
Compre
ssion
-
-F
Shear Force
-φ
+ve V
-ve V
-T
2
Geometric
Property
Material
Properties
4
A – Area (m )
π
Circle A = πR 2 = D 2
4
Rectangle A = bh ,
1
Triangle A = bh
2
J–Polar moment of inertia (m )
E – Young’s Modulus (Pa)
G – Shear Modulus (Pa)
4
Circle: J = πR = πD
2
32
Hollow:
4
π Ro − Ri4
π Do4 − Di4
J=
=
2
32
(
) (
Shear stress:
T
τ= ρ
J
Normal average stress:
F
σ avg =
(Pa)
A
Stresses
4
Uniform
Distribution
)
(Pa)
Shear strain γ - radian
τ = Gγ
Deflection (m) (Elongation)
Angle of Twist (Radian)
L
General: δ =
Deformation
∫
0
F (x )
dx
E (x )A(x )
Shear stress: τ =
Fi Li
External
Work
Castigliano’s
2nd Theorem
Chapters
W =
U =
∫
V
i
Multi-segments: ϕ =
i
1
P∆ p
2
F2
dV =
2 EA
Fi 2 Li
i
∂U
∂P
Chapters 2-3
∆P =
W =
∑ 2E A
i
i
Ti Li
∑G J
i
U=
∫
V
i
1
Tϕ T
2
T2
dV =
2GJ
i
A’
∫
W=
1
1
Pv P or W = M ' θ'
2
2
Ti 2 Li
∂U
∂T
Chapters 4
ϕT =
t
Slope of Elastic Curve (Radian)
dv
M (x )
θ=
=
dx + C
dx
EI
∑ 2G J
i
VQ
(Pa)
It
∫∫
TL
Single segment: ϕ =
GJ
∑E A
i
∫ G(x )J (x )dx
0
FL
Single segment: δ =
EA
Multi-segments: δ =
Total Strain
Energy
General: ϕ =
M
y’
y
V
N.A.
I
Normal strain or Shear strain
σ = Eε or τ = Gγ
Deflection v (m)
(Elastic Curve Equation)
M (x )
v=
dxdx + Cx + D
EI
T (x )
L
y
Q=y’A’
Normal strain: ε = ∆L / L
σ = Eε
My
(Pa)
I
N.A.
T
Strains
Hooke’s Law
bh 3
12
bh 3
Triangle I =
36
E – Young’s Modulus (Pa)
Rectangle I =
Normal stress: σ = −
Distribution of
shear stress
ρ
I–Second moment of area (m4)
πR 4 πD 4
Circle: I =
=
4
64
U=
i i
vP =
∫
V
M2
dV
2 EI
∂U
∂U
or θ M ' =
∂P
∂M '
Chapters 5-8
Lecture Notes of Mechanics of Solids, Chapter 8
14