structural analysis of beams notes.pdf
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Course note on Beam Structural AnalysisTRANSCRIPT
1
DEFLECTION OF BEAMS
Structures undergo deformation when subjected to loads. As a result of this
deformation, deflection and rotation occur in structures. This deformation will disappear
when the loads are removed provided the elastic limit of the material is not exceeded.
Deformation in a structure can also occur due to change in temperature & settlement of
supports.
Deflection in any structure should be less than specified limits for satisfactory
performance. Hence computing deflections is an important aspect of analysis of structures.
There are various methods of computing deflections. Two popular methods are
i) Moment area Method, and
ii) Conjugate beam method
In both of these methods, the geometrical concept is used. These methods are ideal
for statically determinate beams. The methods give a very quick solution when the beam is
symmetrical.
Moment Area Method
This method is based on two theorems which are stated through an example.
Consider a beam AB subjected to some arbitrary load as shown in Figure 1.
Let the flexural rigidity of the beam be EI. Due to the load, there would be bending
moment and BMD would be as shown in Figure 2. The deflected shape of the beam which is
the elastic curve is shown in Figure 3. Let C and D be two points arbitrarily chosen on the
beam. On the elastic curve, tangents are drawn at deflected positions of C and D. The angles
made by these tangents with respect to the horizontal are marked as Cθ and Dθ . These
angles are nothing but slopes. The change is the angle between these two tangents is
demoted as CDθ . This change in the angel is equal to the area of the EI
M diagram between the
two points C and D. This is the area of the shaded portion in figure 2.
Hence CDθ = Cθ Dθ = Area of EI
M diagram between C and D
CDθ = Area BM 1 (a)
EI
It is also expressed in the integration mode as
CDθ = dxCD EI
M 1 (b)
Equation 1 is the first moment area theorem which is stated as follows:
Statement of theorem I:
The change in slope between any two points on the elastic curve for a member
subjected to bending is equal to the area of EI
M diagram between those two points.
2
In figure 4, for the elastic curve a tangent is drawn at point C from which the vertical
intercept to elastic curve at D is measured. This is demoted as KCD. This vertical intercept is
given by
KCD = (Area BM X)CD 2 (a)
EI
Where X is the distance to the centroid of the shaded portion of EI
M diagram
measured from D. The above equation can be expressed in integration mode as
Fig. 1
Fig. 2
Fig. 3
Fig. 4
3
KCD = EI
Mxdx
CD
2 (b)
Equation (2) is the second moment area theorem which is stated as follows.
Statement of theorem II :
The vertical intercept to the elastic curve measured from the tangent drawn to the
elastic curve at some other point is equal to the moment of EI
M diagram, moment being
taken about that point where vertical intercept is drawn.
Sign Convention:
While computing Bending moment at a section, if free body diagram of Left Hand
Portion (LHP) is considered, clockwise moment is taken as positive. If free body diagram of
Right Hand Portion (RHP) is considered, anticlockwise moment is taken as positive. While
sketching the Bending Moment Diagram (BMD), Sagging moment is taken as positive and
Hogging moment is taken as negative.
Proof of Moment Area Theorems:
Figure 5 shows the elastic curve for the elemental length dx of figure 2 to an enlarged
scale. In this figure, R represents the radius of curvature. Then from equation of bending,
with usual notations,
I
M =
R
E (3)
From figure 5,
Rdθ = dx
Hence R = dθ
dx
Substituting this value of R in equation (3),
I
M =
dθ
dx
E
I
M = E
dx
dθ
dθ = EI
M dx
dθ is nothing but change in angle over the elemental length dx. Hence to compute
change in angle from C to D,
θCD = CDdθ =
CDEI
M dx
Hence the proof.
4
C
1 2D
Figure 6 shows the elastic curve from C to D. Change in slope from 1 to 2 is dθ. Distance of
elemental length from D is x.
dΔ = xdθ = x EI
M dx
Therefore, Δ from C to D = CD
EI
M xdx
R
dθ Fig. 5
Fig. 5 Fig. 6
KCD
dθ
d
dθ
5
Problem 1 : Compute deflections and slopes at C,D and E. Also compute slopes at A and B.
To Compute Reactions:
00fxA
0WWVV0fyB
A
2WVV BA
+
03
2LW
3
WLLV0M
AB
6
WL3
2WL
3
WLLV
A
VA = W ; VB =W
Bending Moment Calculations:
Section (1) – (1) (LHP, 0 to L/3)
+ Mx-x = Wx
At x = 0; BM at A = 0
x = 3L ; BM @ C = 3
WL
Section (2) – (2) (LHP, 3L to 3
2L )
+ Mx-x = Wx – W(x - 3L )
At x = 3L , BM @ C = 3
L3
L3
L WWW
= 3LW
At x = 3
2L, BM @ D =
3
L
3
2LW
3
2LW
=
3
WL
3
2WL
3
2WL
= 3
WL
Section (3) – (3) RHP (0 to 3L )
+ Mx-x = Wx
At x = 0; BM @ B = 0
At x = 3L , BM @ D =
3
WL
This beam is symmetrical. Hence the BMD & elastic curve is also symmetrical. In
such a case, maximum deflection occurs at mid span, marked as δE. Thus, the tangent drawn
at E will be parallel to the beam line and θE is zero.
Also, δc = δD, θA = θB and θC = θD
To compute θC
From first theorem,
θCE = Area of BMD between E&C
EI
θC~ θE =
EI
W 6L
3L
= 18EI
WL2
θE being zero, θC = WL2
(
)
18EI
7
To compute θΔ
From First theorem,
θΔE = Area of BMD between A&E
EI
θA~ θE = EI
6
L
3
WL
3
WL
3
L2
1
= EI
18
WL
18
WL 22
θE being zero, θA = 9EI
WL2
( )
θB = 9EI
WL2
( )
To compute δE
From 2nd
theorem
KEA =
EI
X BM of AreaEA
KEA =
EI
12
L
3
L
6
L
3
WL
3
L
3
2
3
WL
3
L
2
1
=
EI
216
5WL
81
WL 33
=
648
15WL8WL
EI
1 33
= 648EI
23WL3
8
From figure, KEA is equal to δE.
Therefore δE = 648EI
23WL3
To compute θC
From 2nd
theorem
KEC =
EI
X BMD of AreaCE
=
EI
W 12L
6L
3L
=
216
1
EI
WL3
= 216EI
WL3
δc = δE - KEC
216EI
WL
648EI
23WLδ
33
C
= 648EI
3WL23WL 33
= 648EI
20WL3
= 162EI
5WL3
= 162EI
5WLδδ
3
CD
9
Problem 2. For the cantilever beam shows in figure, compute deflection and slope at the
free end.
Consider a section x-x at a distance x from the free end. The FBD of RHP is taken into
account.
(RHP +) BM @ X-X = MX-X = -10 (x) (x/2) = -5x2
At x = 0; BM @ B = 0
At x = 4m; BM @ A = -5(16) = -80 kNm
The BMD is sketched as shown in figure. Note that it is Hogging Bending Moment.
The elastic curve is sketched as shown in figure.
10
To compute θB
For the cantilever beam, at the fixed support, there will be no rotation and hence in
this case θA = 0. This implies that the tangent drawn to the elastic curve at A will be the same
as the beam line.
From I theorem,
θAB = θA ~ θB = 4
0EI
Mdx
= dx5XEI
14
0
2
= 403x3
EI
5
= 3EI
32064
3EI
5
θA being zero,
θB = 3EI
320 ( )
To compute δB
From II theorem
KAB = 4
0EI
Mxdx
= xdx5XEI
14
0
2
= 2564EI
5
EI
5 4
04x4
= EI
320
From the elastic curve,
KAB = δB = EI
320
11
Problem 3: Find deflection and slope at the free end for the beam shown in figure by using
moment area theorems. Take EI = 40000 KNm-2
Calculations of Bending Moment:
Region AC: Taking RHP +
Moment at section = -6x2/2
= -3x2
At x = 0, BM @ A = 0
x = 4m; BM @ C = -3(16) = - 48kNm
12
Region CB: (x = 4 to x = 8)
Taking RHP +, moment @ section = -24 (x-2)
= -24x+48;
At x = 4m; BM @ C = -24(4) + 48 = -48kNm;
x = 8 m BM @ B = -144 kNm;
To compute θB:
First moment area theorem is used. For the elastic curve shown in figure. We know
that θA = 0.
θAB = θA ~ θB = EI
Mdx
= 8
4
4
0
2 dx4824xEI
1dx3x
EI
1
842x
4
03x
A48x24
EI
1
EI
3θ
23
= 4848166412EI
1
EI
64
= -0.0112 Radians
= 0.0112 Radians ( )
To compute δB
EI
MxdxKAB
=
8
4
4
0
2 xdx4824xEI
1xdx3x
EI
1
= 8
42x
8
43x
4
04x 234
4824EI
1
EI
3
=
16642464512
3
24
EI
1256
4EI
3
= 11523584EI
1
EI
192
=
0.0656m0.0656mEI
2624
13
Problem 4: For the cantilever shown in figure, compute deflection and at the points where
they are loaded.
To compute θB :
θBA = θB ~ θA = 151.537.52.5EI
12
12
1
θB = EI
58.125 ( )
θC = 151.51537.51.5EI
12
12
1
14
= EI
50.625 ( )
δB =
1451.5EI
1
EI
2.537.52.52
13
22
1
= EI
100.625
= EI
100.625
δC = 1451.50.8571537.51.5EI
12
12
1
δC = EI
44.99
STRAIN ENERGY 4.1 Introduction
Under action of gradually increasing external loads, the joints of a structure deflect
and the member deform. The applied load produce work at the joints to which they are
applied and this work is stored in the structure in the form of energy known as Strain
Energy. If the material of structure is elastic, then gradual unloading of the structure relieves
all the stresses and strain energy is recovered.
The slopes and deflections produced in a structure depend upon the strains developed
as a result of external actions. Strains may be axial, shear, flexural or torsion. Therefore, ther
is a relationship can be used to determine the slopes and deflections in a structure.
4.2 Strain energy and complementary strain energy
When external loads are applied to a skeletal structure, the members develop internal
force „F‟ in the form of axial forces („P‟), shear force („V‟) , bending moment (M) and
twisting moment (T). The internal for „F‟ produce displacements „e‟. While under goint these
displacements, the internal force do internal work called as Strain Energy
Figure 1 shows the force displacement relationship in which Fj is the internal force
and ej is the corresponding displacement for the jth element or member of the structure.
15
The element of internal work or strain energy represented by the area the strip with
horizontal shading is expressed as:
Strain energy stored in the jth
element represted by the are under force-displacement
curve computed as :
For m members in a structure, the total strain energy is
The area above the force-displacement curve is called Complementary Energy. For
jth
element, the complementary strain energy is represented by the area of the strip with
vertical shading in Fig.1 and expressed as
Complementary strain energy of the entire structure is
Complementary strain energy of the entire structure is
Fj
Strain Energy(Ui)j
ej ej ej+ej
Fj
Fj+Fj
Complementry SE(Ui)j
Fig.1 FORCE-DISPLACEMENT RELATIONSHIP
.....(1) eF U jji
.....(2) deF )U( jjji
.....(3) deF )(U Um
1j
m
1j
jjjii
.....(4) Fe U jji
.....(5) dFe )U( jjji
.....(6) dFe Um
1j
jji
16
When the force-displacement relationship is linear, then strain energy and
complimentary energies are equal
4.3 Strain energy expressions
Expression for strain energy due to axial force, shear force and bending moment is
provided in this section
4.3.1 Strain energy due to Axial force
A straight bar of length „L‟ , having uniform cross sectional area A and E is the Young‟s
modulus of elasticity is subjected to gradually applied load P as shown in Fig. 2. The bar
deforms by dL due to average force 0+(P/2) = P/2. Substituting Fj = P/2 and dej = dl in
equation 2, the strain energy in a member due to axial force is expressed as
From Hooke‟s Law, strain is expressed as
Hence
L A,E
dL
.....(7) UU ii
(8) ....... dL 2
P )(U Pi
A
P where,
E
dx
dL
...(9).......... AE
PLx dL
Fig.2
17
Substituting equation 9 in 8, strain energy can be expressed as
For uniform cross section strain energy expression in equation 10 can be modified as
If P, A or E are not constant along the length of the bar, then equation 10 is used instead of
10a.
4.3.1 Strain energy due to Shear force
A small element shown in Fig.3 of dimension dx and dy is subjected to shear force Vx . Shear
stress condition is shown in Fig. 4. Shear strain in the element is expressed as
Where, Ar= Reduced cross sectional area and G= shear modulus
Shear deformation of element is expressed as
Substituting Fj = Vx/2, dej = dev in equation (2) strain energy is expressed as
dy dx dy
dx
(10) ....... 2AE
dxP )(U
L
0
2
Pi
a) (10 ....... 2AE
LP )(U
2
Pi
Fig.3 Fig.4
......(11) GA
V
r
x
...(12).......... GA
dx V de
r
xv
(13) ....... G2A
dxV )(U
L
0 r
2
xVi
18
4.3.2 Strain energy due to Bending Moment
An element of length dx of a beam is subjected to uniform bending moment „M‟. Application
of this moment causes a change in slope d is expressed as
Where , EI
M
R
1 x , Substituting Fj = Mx/2, dej= deM in equation (2), Strain energy due to
bending moment is expressed as
4.3 Theorem of minimum Potential Energy
Potential energy is the capacity to do work due to the position of body. A body of weight
„W‟ held at a height „h‟ possess an energy „Wh‟. Theorem of minimum potential energy
states that “ Of all the displacements which satisfy the boundary conditions of a
structural system, those corresponding to stable equilibrium configuration make the
total potential energy a relative minimum”. This theorem can be used to determine the
critical forces causing instability of the structure.
4.3 Law of Conservation of Energy
From physics this law is stated as “Energy is neither created nor destroyed”. For the purpose
of structural analysis, the law can be stated as “ If a structure and external loads acting on
it are isolated, such that it neither receive nor give out energy, then the total energy of
the system remain constant”. With reference to figure 2, internal energy is expressed as in
equation (9). External work done We = -0.5 P dL. From law of conservation of energy Ui+We
=0. From this it is clear that internal energy is equal to external work done.
4.3 Principle of Virtual Work:
Virtual work is the imaginary work done by the true forces moving through imaginary
displacements or vice versa. Real work is due to true forces moving through true
......(14) EI
dxM
R
dx dde x
M θ
(15) ....... 2EI
dxM )(UL
0
2
xMi
19
displacements. According to principle of virtual work “ The total virtual work done by a
system of forces during a virtual displacement is zero”.
Theorem of principle of virtual work can be stated as “If a body is in equilibrium under a
Virtual force system and remains in equilibrium while it is subjected to a small
deformation, the virtual work done by the external forces is equal to the virtual work
done by the internal stresses due to these forces”. Use of this theorem for computation of
displacement is explained by considering a simply supported bea AB, of span L, subjected to
concentrated load P at C, as shown in Fig.6a. To compute deflection at D, a virtual load P‟ is
applied at D after removing P at C. Work done is zero a s the load is virtual. The load P is
then applied at C, causing deflection C at C and D at D, as shown in Fig. 6b. External work
done We by virtual load P‟ is . If the virtual load P‟ produces bending moment
M‟, then the internal strain energy stored by M‟ acting on the real deformation d in element
dx over the beam equation (14)
Where, M= bending moment due to real load P. From principle of conservation of energy
We=Wi
2
δP' W D
e
L
0i
U
0
L
0i
EI 2
dx MM' U;
2
dθM' dU
L
0
D
EI 2
dx MM'
2
δP'
20
If P‟=1 then
Similarly for deflection in axial loaded trusses it can be shown that
Where,
= Deflection in the direction of unit load
P‟ = Force in the ith
member of truss due to unit load
P = Force in the ith
member of truss due to real external load
n = Number of truss members
L = length of ith
truss members.
Use of virtual load P‟ = 1 in virtual work theorem for computing displacement is called
Unit Load Method
4.4 Castigliano‟s Theorems:
Castigliano published two theorems in 1879 to determine deflections in structures and
redundant in statically indeterminate structures. These theorems are stated as:
A B C D
a
x L
P
A B C D
a
x L
P P’ C D
Fig.6a
Fig.6b
(16) EI
dx MM' δ L
0D
(17) AE
dx PP' δ
n
0
21
1st Theorem : “If a linearly elastic structure is subjected to a set of loads, the
partial derivatives of total strain energy with respect to the deflection at any point is
equal to the load applied at that point”
2nd Theorem: “If a linearly elastic structure is subjected to a set of loads, the
partial derivatives of total strain energy with respect to a load applied at any point is
equal to the deflection at that point”
The first theorem is useful in determining the forces at certain chosen coordinates. The
conditions of equilibrium of these chosen forces may then be used for the analysis of
statically determinate or indeterminate structures. Second theorem is useful in computing the
displacements in statically determinate or indeterminate structures.
4.5 Betti‟s Law:
It states that If a structure is acted upon by two force systems I and II, in equilibrium
separately, the external virtual work done by a system of forces II during the
deformations caused by another system of forces I is equal to external work done by I
system during the deformations caused by the II system
A body subjected to two system of forces is shown in Fig 7. Wij represents work done by ith
system of force on displacements caused by jth
system at the same point. Betti‟s law can be
(18) N ..... 1,2, j P δ
Uj
j
(19) N .1,2,......j δ P
Uj
j
I II
Fig. 7
22
expressed as Wij = Wji, where Wji represents the work done by jth
system on displacement
caused by ith
system at the same point.
Numerical Examples
1. Derive an expression for strain energy due to bending of a cantilever beam of length
L, carrying uniformly distributed load „w‟ and EI is constant
Solution:
Bending moment at section 1-1 is
Strain energy due to bending is
w
x
1
1
2
wx- M
2
x
2EI
dxM )(UL
0
2
xMi
L
0
L
0
5242L
0
22
i40EI
xwdx
8EI
xw
2EI
dx2
wx-
U
Answer 40EI
Lw U
52
i
23
2. Compare the strain energies due to three types of internal forces in the rectangular
bent shown in Fig. having uniform cross section shown in the same Fig. Take E=2 x 105
MPa, G= 0.8 x 105 MPa, Ar= 2736 mm2
Solution:
Step 1: Properties
A=120 * 240 – 108 * 216 = 5472 mm2,
E= 2 * 105 MPa ; G= 0.8 * 105 MPa ; Ar = 2736 mm2
Step 2: Strain Energy due to Axial Forces
Member AB is subjected to an axial comprn.=-12 kN
Strain Energy due to axial load for the whole str. is
Step 3: Strain Energy due to Shear Forces
Shear force in AB = 0; Shear force in BC = 12 kN
Strain Energy due to Shear for the whole str. Is
Step 4: Strain Energy due to Bending Moment
Bending Moment in AB = -12 * 4 = -48 kN-m
Bending Moment in BC = -12 x
Strain Energy due to BM for the whole structure is
120 mm
240 mm 12 mm 5m
4m
12kN
A
B C
4633
mm 10 x 47.54 12
216 * 108 -
12
240 * 120 I
mm-N 328.94 10*2 * 5472 * 2
5000 * )10 * (-12
2AE
LP )(U
5
232n
1i
2
Pi
mm-N 78.1315 x100.8 * 2736 * 2
4000 * )10 * (12
G2A
LV )(U
5
232n
1ir
2
xVi
24
Step 5: Comparison
Total Strain Energy = (Ui)p + (Ui)V+ (Ui)M
Total Strain Energy =328.94 +1315.78 +767.34 x 103
= 768.98 x 103 N-mm
Strain Energy due to axial force, shear force and bending moment are 0.043%, 0.17% &
99.78 % of the total strain energy.
3. Show that the flexural strain energy of a prismatic bar of length L bent into a complete
circle by means of end couples is
Solution:
Circumference = 2 R =L or
From bending theory
M M
L
R
10* 47.54 * 2x10 * 2
5000 * )10 * (-48
2EI
dxM )(U
65
262n
1i
2
xMi
mm-N 10*767.34 10* 47.54 * 10*2 * 2
dxx)*10 *(-12 34000
065
23
2L
πEI2
applied couple M whereL
EI2
2πL
EIM
Answer
L
EI2π
2EI
LL
πEI2
2EI
LM )(U
22
2
Mi
25
4. Calculate the strain energy in a truss shown in Fig. if all members are of same cross-
sectional area equal to 0.01m2 and E=200GPa
Solution: To calculate strain energy of the truss, first the member forces due to external force
is required to be computed. Method of joint has been used here to compute member forces.
Member forces in the members AB, BC, BD, BE, CE and DE are only computed as the truss
is symmetrical about centre vertical axis.
Step1: Member Forces:
i) Joint A: From triangle ACB, the angle = tan-1(3/4)=36052‟
The forces acting at the joint is shown in Fig. and the forces in members are computed
considering equilibrium condition at joint A
Fy=0; FABsin+30=0; FAB=-50kN (Compression)
Fx=0; FABcos + FAC=0; FAC=40kN (Tension)
FAC
FAB
RA= 30 kN
4m
AGC E 4m4m4m
30 kN
B D
30 kN
3m
F
H
26
ii) Joint C: The forces acting at the joint is shown in Fig. and the forces in members
are computed considering equilibrium condition at joint C
Fy=0; FCB=0;
Fx=0; FCE - 40=0; FCE=40kN(Tension)
iii) Joint B: The forces acting at the joint is shown in Fig. and the forces in members
are computed considering equilibrium condition at joint B
Fy=0; -30+50 sin-FBEsin=0; FBE=0
Fx=0; 50 cos - FBD=0; FBD=-40kN (Compression)
iv) Joint D: The forces acting at the joint is shown in Fig. and the forces in members
are computed considering equilibrium condition at joint D
Fy=0; FDE=0; Fx=0; FDF + 40=0; FDF=-40kN (Comprn.)
Forces in all the members are shown in Fig.
FCE FAC=40kN
FCB
FBD
30kN
FAB= 50kN FCB=0
FBE
FDF FBD=40 FDE
27
00000-50
40 40 40 40
-50
-40-40
H
FDB
EC GA
Step 2: Strain Energy
A= 0.01m2; E=2*105 N/mm2 = 2*108 kN/m2, AE = 2*106 kN
(Ui)p=15.83*10-3
kN-m
5. Determine the maximum slope and maximum deflection in a cantilever beam of span L
subjected to point load W at its free end by using strain energy method. EI is constant
Solution:
i) Maximum Deflection
BM at 1-1 Mx= -Wx
From 2nd theorem of Castigliaino
L
x
1
1 W
A B
4*(-40)*25)50(24*40*4 10*2 * 2
1
2AE
LP )(U 222
6
13n
1i
2
Pi
B
L
0
xx
δ EI
dx M W
M
M
U
L
0B
EI
dx (-Wx) (-x)δ x,-
W
M
Answer 3EI
WL
3
x
EI
W dx x
EI
Wδ
3L
0
L
0
32
B
28
ii) Maximum Slope
Maximum slope occurs at B, Virtual moment M‟ is applied at B
Bending moment at 1-1 is Mx = -Wx – M‟
From 2nd theorem of Castiglano
Substituting M‟=0
6. Calculate max slope and max deflection of a simply supported beam carrying udl of
intensity w per unit length throughout its length by using Castigliano‟s Theorem
L
x
1
1 W
A M’ B
L
w
EI
dx M M'
M
M'
Uθ
L
0
xx
B
L
0B
x
EI
dx )M' -(-Wx (-1)θ 1,-
M'
M
Answer 2EI
WL
2
x
EI
W dx x
EI
Wθ
2L
0
L
0
2
B
29
i) Maximum Slope:
Maximum slope occurs at support. A virtual moment M‟ is applied at A.
Reactions:
BM at 1-1
Put M‟=0
ii) Maximum Deflection:
Maximum Deflection occurs at mid span. A virtual downward load W‟ is applied at
mid-span.
L
w
1
1
x
M’
RA RB
L
M'
2
wLR ;
L
M' -
2
wLR BA
L
x-1
M'
M and M'
2
wx - x
L
M' -
2
wLM x
2
x
L
0
2
A dxL
x-1 M'
2
wx - x )
L
M'-
2
wL(
EI
1
M'
Uθ
L
0
32
A dx 2L
wx
2
2wx -x
2
wL(
EI
1θ
L
0
432
A4L
x
3
x -x
2
Lx(
2EI
wθ
Answer 24EI
wLθ
3
A
30
Reactions:
BM at 1-1
Put W‟=0
L
w
1
1
x
M’
RA RB
L/2
2
W'
2
wLR ;
2
W'
2
wLR BA
ACRegion for 2
x
W'
M and
2
wx - x
2
W'
2
wLM x
2
x
L/2
0
2
C dx2
x
2
wx - x )
L
W'
2
wL(
EI
2
W'
U
L/2
0
43L/2
0
32
C4
x -
3
Lx
2EI
w dx x-(Lx
4EI
w2δ
64
L -
24
L
2EI
w δ
L/2
0
44
C
Answer 384EI
5wL δ
4
Cmax
31
CONJUGATE BEAM METHOD
This is another elegant method for computing deflections and slopes in beams. The
principle of the method lies in calculating BM and SF in an imaginary beam called as
Conjugate Beam which is loaded with M/EI diagram obtained for real beam. Conjugate
Beam is nothing but an imaginary beam which is of the same span as the real beam carrying
M/EI diagram of real beam as the load. The SF and BM at any section in the conjugate beam
will represent the rotation and deflection at that section in the real beam. Following are the
concepts to be used while preparing the Conjugate beam.
It is of the same span as the real beam.
The support conditions of Conjugate beam are decided as follows:
Some examples of real and conjugate equivalents are shown.
32
Problem 1 : For the Cantilever beam shown in figure, compute deflection and rotation at
(i) the free end (ii) under the
load
Conjugate Beam:
By taking a section @ C´ and considering FBD of LHP,
EI
2253
EI
150-f SF 2
1x
BM @ C´= ;EI
45023
EI
150-2
1
33
Similarly by taking a section at A‟ and considering FBD of LHP;
SF @ A‟ = EI
225
BM @ A‟ = EI
90022
EI
225
SF @ a section in Conjugate Beam gives rotation at the same section in Real Beam
BM @ a section in Conjugate Beam gives deflection at the same section in Real Beam
Therefore, Rotation @ C = EI
225 ( )
Deflection @ C= EI
450
Rotation @ A = EI
225 ( )
Deflection @ A = EI
900
Problem 2: For the beam shown in figure, compute deflections under the loaded points.
Also compute the maximum deflection. Compute, also the slopes at supports.
34
Note that the given beam is symmetrical. Hence, all the diagrams for this beam should be
symmetrical. Thus the reactions are equal & maximum deflection occurs at the mid span. The
bending moment for the beam is as shown above. The conjugate beam is formed and it is
shown above.
For the conjugate beam:
]Beam Conjugateon load [Total VV 21'
B
'
A
= EI30
EI60
21
21 432
= EI
150
EI
120
EI
1802
1
To compute δC :
A section at C‟ is placed on conjugate beam. Then considering FBD of LHP;
35
+ BM @ C‟= 1EI
6033
EI
1502
1
= EI
360
EI
90
EI
450
; EI
360δ
C
δD = δC (Symmetry)
To compute δE:
A section @ E‟ is placed on conjugate beam. Then considering FBD of LHP;
+ BM @ E‟= 12EI
303
EI
6035
EI
1502
1
i.e δE = EI
420
EI
60
EI
270
EI
750
θA = EI
150 ( ) θB =
EI
150 ( )
36
Problem 3: Compute deflection and slope at the loaded point for the beam shown in figure.
Given E = 210 Gpa and I = 120 x 106mm
4. Also calculate slopes at A and B.
Note that the reactions are equal. The BMD is as shown above.
To Compute reactions in Conjugate Beam:
03
EI
120
2
13
EI
60
2
1VV0fy '
B
'
A
37
EI
720
EI
360
EI
3606V '
A
;
SF and BM at C‟ is obtained by placing a section at C‟ in the conjugate beam.
SF @ C‟ =
+ BM @ C‟ =
Given E = 210 x 109 N/m
2
= 210 x 106 kN/m
2
I = 120 x 106 mm
4
= 120 x 106 (10
-3 m)
4
= 120 x 106 (10
-12)
= 120 x 10-6
m4;
EI = 210 x 106 (120 x 10
-6) = 25200 kNm
-2
Rotation @ C = 25200
30 = 1.19 x 10
-3 Radians ( )
Deflection @ C = 25200
270 = 0.0107 m
= 10.71 mm ( )
θA = 4.76 X 10-3
Radians
θB = 5.95 X 10-3
Radians:
0;EI
180
EI
90VV '
B
'
A
EI
270VV '
B
'
A
023EI
120
2
143
EI
60
2
16V 0m '
AB'
EI
120V '
A
EI
150V '
B
3EI
60
2
1
EI
120
EI
30
13EI
60
2
13
EI
120
EI
270
EI
90
EI
360
38
Problem 4: Compute slopes at supports and deflections under loaded points for the beam
shown in figure.
39
To compute reactions and BM in real beam:
+ 0MB 031006509VA
66.67kN9
600VA 83.33kNVB
BM at (1) – (1) = 66.67 x
At x = 0; BM at A = 0, At x = 3m, BM at C = 200 kNm
BM at (2) – (2) = 66.67 x – 50 (x-3) = 16.67 x + 150
At x = 3m; BM at C = 200 kNm, At x = 6m, BM at D = 250 kNm
BM at (3) – (3) is computed by taking FBD of RHP. Then
BM at (3)-(3) = 83.33 x (x is measured from B)
At x = 0, BM at B = 0, At x = 3m, BM at D = 250 kNm
To compute reactions in conjugate beam:
+ 0M '
B
i.e 02EI
83.333
2
14
EI
253
2
14.5
EI
10037
EI
2003
2
19V '
A
EI
38509V '
A
( ) ( )
EI
1003
EI
2003
2
1VV0fy '
B
'
A
EI
83.333
2
1
EI
253
2
1
EI
762.5
EI
427.77V '
A
EI
334.73V '
B
EI
427.77θ
A EI
334.73θ
B
150VV0fy BA
40
To Compute δC :
A Section at C‟ is chosen in the conjugate beam:
+ BM at C‟ = 1
EI
2003
2
13
EI
427.77
= EI
983.31
δC = EI
983.31
To compute δD:
Section at D‟ is chosen and FBD of RHP is considered.
+ BM at D‟ = 1EI
83.333
2
13
EI
334.73
= EI
879.19
EI
879.19δ
D
41
Problem 5: Compute to the slope and deflection at the free end for the beam shown in
figure.
42
The Bending moment for the real beam is as shown in the figure. The conjugate beam also is
as shown.
Section at A‟ in the conjugate beam gives
SF @ A‟ = 4
0
2
dxEI
5x
= 643EI
5
EI
5 4
03x3
= 3EI
320
θA = 3EI
320 ( )
BM @ A‟ = dxx5xEI
1 4
0
2
= 2564EI
5
4
x
EI
5-4
0
4
δA = EI
320
43
STRUCTURAL ANALYSIS-I (CV42)
CHAPTER 1
1.1 Introduction
A structure can be defined as a body which can resist the applied loads without
appreciable deformations. Civil engineering structures are created to serve some specific
functions like human habitation, transportation, bridges, storage etc. in a safe and economical
way. A structure is an assemblage of individual elements like pin ended elements (truss
elements), beam element, column, shear wall slab cable or arch. Structural engineering is
concerned with the planning, designing and the construction of structures. Structural analysis
involves the determination of the forces and displacements of the structures or components of
a structure. Design process involves the selection and or detailing of the components that
make up the structural system. The cyclic process of analysis and design is illustrated in the
flow chart given in Fig.1.1
1.2 Forms of Structures
Engineering structure is an assemblage of individual members. Assemblage of members
forming a frame to support the forces acting is called framed structure. Assemblage of
No
Preliminary
Design
Structural
Analysis
Compute
Stresses and
deformations
Is stresses and
Deformation
within limits?
Final Design
Revise
sections
Yes
Fig.1.1 Cyclic Process
44
continuous members like flat plates, curved members etc., are called continuous system.
Buildings, bridges, transmission towers, space crafts, aircrafts etc., are idealized as framed
structures. Shells, domes, plates, retaining walls, dams, cooling towers etc., are idealized as
continuous systems.
A frame work is the skeleton of the complete structure. This frame work supports all
intended loads safely and economically. Continuous system structures transfers loads through
the in-plane or membrane action to the boundaries. Fig.1.2 and Fig.1.3 illustrates framed
structure and continuous system.
Actual structure is generally converted to simple single line structures and this process
is called idealization of structures. The idealization consists of identifying the members of
structure as well known individual structural elements. This process requires considerable
experience and judgment. Structural analyst may be required to idealize the structure as one
or more of following
Fig.1.2 Framed Structure
45
Fig.1.3 Continuous system
46
i) Real structure ii) A physical model iii) A mathematical model
In a real structure the response of the structure is studied under the actual forces like
gravity loads and lateral loads. The load test is performed using elaborate loading equipment.
Strains and deformations of structural elements under loads are measured. This is very
expensive and time consuming procedure, hence performed in only exceptional cases. Load
testing carried out on a slab system is illustrated in Fig. 1.4
Fig.1.4 Load test on slab
47
Physical models which are scaled down and made up of plastic, metal or other
suitable materials are used to study the response of structure under loading. These models are
tested in laboratories. This study requires special techniques and is expensive. This study is
carried out under compelling circumstances. Examples includes laboratory testing of small
scale building frames, shake table test of bridges and building, photo elastic testing of a dam
model, wind tunnel testing of small scale models of high rise buildings, towers or chimneys.
Fig. 1.5 shows testing of a slab model under uniformly distributed load.
A mathematical model is the development of mathematical equations. These
equations describe the structure loads and connections. Equations are then solved using
suitable algorithm. These solutions generally require electronic computers. The process of
mathematical modeling is shown in block diagram given in Fig.1.6
A structure is generally idealized as either two dimensional structure (Plane frame) or
as three dimensional structure (Space frame). The selection of idealization depends on the
desire and experience of structural engineer. A two dimensional structure or a plane frame
structure is that which has all members and forces are in one plane. Space frame or a three
dimensional structure has members and forces in different planes. All structures in practice
are three dimensional structures. However, analyst finds more convenient to analyze a plane
Fig.1.5 Testing of a Model
48
structure rather than a space structure. Fig. 1.7 shows two dimensional and three dimensional
structures used in mathematical modeling.
Fig.1.5 b. Three Dime
A mathematical modeling should also idealize the supports of the structure. Roller
supports or simple supports, pinned supports or hinged supports and fixed supports are
generally assumed type of supports in practice. Fig. 1.8 shows different types of supports. In
a roller support the reaction is perpendicular to the surface of the roller. Two components of
Actual Structure
Idealize Structure
Idealize Loads
Development of
Equations
Response of str.
Interpretation of
Results
Fig.1.6 Block Diagram of Mathematical modeling
Fig.1.7 a. Two Dimensional Structures
Fig.1.7 b. Three Dimensional Structures
49
reaction are developed in hinged support and three reaction component, one moment and two
forces parallel to horizontal and vertical axis are developed in fixed support.
Mathematical modeling requires to consider the loads acting on structure. Determination
of the loads acting on the structure is often difficult task. Minimum loading guidance exists
in codes and standards. Bureau of Indian standards, Indian road congress and Indian railways
have published loading standards for building, for roads and for railway bridges respectively.
Loads are generally modeled as concentrated point loads, line loads or surface loads. Loads
are divided into two groups viz., dead loads and live loads. Dead loads are the weight of
structural members, where as live loads are the forces that are not fixed. Snow loads, Wind
loads, Occupancy loads, Moving vehicular loads, Earth quake loads, Hydrostatic pressure,
earth pressure , temperature and fabrication errors are the live loads. All the live loads may
not act on the structure simultaneously. Judgment of analyst on this matter is essential to
avoid high loads.
Fig.1.8 Typical Support Conditions a) Fixed support, b) Hinged support, c) simple support
50
1.3 Conditions of Equilibrium and Static Indeterminacy
A body is said to be under static equilibrium, when it continues to be under rest after
application of loads. During motion, the equilibrium condition is called dynamic
equilibrium. In two dimensional system, a body is in equilibrium when it satisfies following
equation.
Fx=0 ; Fy=0 ; Mo=0 ---1.1
To use the equation 1.1, the force components along x and y axes are considered. In
three dimensional system equilibrium equations of equilibrium are
Fx=0 ; Fy=0 ; Fz=0;
Mx=0 ; My=0 ; Mz=0; ----1.2
To use the equations of equilibrium (1.1 or 1.2), a free body diagram of the structure
as a whole or of any part of the structure is drawn. Known forces and unknown reactions with
assumed direction is shown on the sketch while drawing free body diagram. Unknown forces
are computed using either equation 1.1 or 1.2
Before analyzing a structure, the analyst must ascertain whether the reactions can be
computed using equations of equilibrium alone. If all unknown reactions can be uniquely
determined from the simultaneous solution of the equations of static equilibrium, the
reactions of the structure are referred to as statically determinate. If they cannot be
determined using equations of equilibrium alone then such structures are called statically
indeterminate structures. If the number of unknown reactions are less than the number of
equations of equilibrium then the structure is statically unstable.
The degree of indeterminacy is always defined as the difference between the number
of unknown forces and the number of equilibrium equations available to solve for the
unknowns. These extra forces are called redundants. Indeterminacy with respect external
forces and reactions are called externally indeterminate and that with respect to internal
forces are called internally indeterminate.
A general procedure for determining the degree of indeterminacy of two-dimensional
structures are given below:
NUK= Number of unknown forces
NEQ= Number of equations available
IND= Degree of indeterminacy
IND= NUK - NEQ
51
Indeterminacy of Planar Frames
For entire structure to be in equilibrium, each member and each joint must be in
equilibrium (Fig. 1.9)
NEQ = 3NM+3NJ
NUK= 6NM+NR
IND= NUK – NEQ = (6NM+NR)-(3NM+3NJ)
IND= 3NM+NR-3NJ ----- 1.3
Degree of Indeterminacy is reduced due to introduction of internal hinge
NC= Number of additional conditions
NEQ = 3NM+3NJ+NC
NUK= 6NM+NR
IND= NUK-NEQ = 3NM+NR-3NJ-NC ------------1.3a
Indeterminacy of Planar Trusses
Members carry only axial forces
NEQ = 2NJ
NUK= NM+NR
IND= NUK – NEQ
IND= NM+NR-2NJ ----- 1.4
Three independent reaction components
Two independent reaction components
Fig. 1.9 Free body diagram of Members and Joints
52
Indeterminacy of 3D FRAMES
A member or a joint has to satisfy 6 equations of equilibrium
NEQ = 6NM + 6NJ-NC
NUK= 12NM+NR
IND= NUK – NEQ
IND= 6NM+NR-6NJ-NC ----- 1.5
Indeterminacy of 3D Trusses
A joint has to satisfy 3 equations of equilibrium
NEQ = 3NJ
NUK= NM+NR
IND= NUK – NEQ
IND= NM+NR-3NJ ----- 1.6
Stable Structure:
Another condition that leads to a singular set of equations arises when the body or
structure is improperly restrained against motion. In some instances, there may be an
adequate number of support constraints, but their arrangement may be such that they
cannot resist motion due to applied load. Such situation leads to instability of structure. A
structure may be considered as externally stable and internally stable.
Externally Stable:
Supports prevents large displacements
No. of reactions ≥ No. of equations
Internally Stable:
Geometry of the structure does not change appreciably
For a 2D truss NM ≥ 2Nj -3 (NR ≥ 3)
For a 3D truss NM ≥ 3Nj -6 (NR ≥ 3)
f)
53
Examples:
1.1 Determine Degrees of Statical indeterminacy and classify the structures
a)
b)
c)
d)
e)
f)
NM=2; NJ=3; NR =4; NC=0
IND=3NM+NR-3NJ-NC
IND=3 x 2 + 4 – 3 x 3 -0 = 1
INDETERMINATE
NM=3; NJ=4; NR =5; NC=2
IND=3NM+NR-3NJ-NC
IND=3 x 3 + 5 – 3 x 4 -2 = 0
DETERMINATE
NM=3; NJ=4; NR =5; NC=2
IND=3NM+NR-3NJ-NC
IND=3 x 3 + 5 – 3 x 4 -2 = 0
DETERMINATE
NM=3; NJ=4; NR =3; NC=0
IND=3NM+NR-3NJ-NC
IND=3 x 3 + 3 – 3 x 4 -0 = 0
DETERMINATE
NM=1; NJ=2; NR =6; NC=2
IND=3NM+NR-3NJ-NC
IND=3 x 1 + 6 – 3 x 2 -2 = 1
INDETERMINATE
NM=1; NJ=2; NR =5; NC=1
IND=3NM+NR-3NJ-NC
IND=3 x 1 + 5 – 3 x 2 -1 = 1
INDETERMINATE
54
R2
R1
R5
R4
g)
NM=8; NJ=8; NR =24; NC=0
IND=6NM+NR-6NJ-NC
IND=6 x 8 + 24 – 6 x 8 -0 = 24
INDETERMINATE
Each support has 6 reactions
NM=18; NJ=15; NR =18; NC=0
IND=6NM+NR-6NJ-NC
IND=6 x 18 + 18 – 6 x 15 = 36
INDETERMINATE
Each support has 3 reactions
NM=1; NJ=2; NR =5; NC=1
IND=3NM+NR-3NJ-NC
IND=3 x 1 + 5 – 3 x 2 -1 = 1
INDETERMINATE
h)
i)
Truss
NM=2; NJ=3; NR =4;
IND=NM+NR-2NJ
IND= 2 + 4 – 2 x 3 = 0
DETERMINATE
j)
55
1.4 Degree of freedom or Kinematic Indeterminacy
Members of structure deform due to external loads. The minimum number of parameters
required to uniquely describe the deformed shape of structure is called “Degree of
Freedom”. Displacements and rotations at various points in structure are the parameters
considered in describing the deformed shape of a structure. In framed structure the
deformation at joints is first computed and then shape of deformed structure.
Deformation at intermediate points on the structure is expressed in terms of end
deformations. At supports the deformations corresponding to a reaction is zero. For
example hinged support of a two dimensional system permits only rotation and
translation along x and y directions are zero. Degree of freedom of a structure is
expressed as a number equal to number of free displacements at all joints. For a two
dimensional structure each rigid joint has three displacements as shown in Fig. 1.10
In case of three dimensional structure each rigid joint has six
displacement.
• Expression for degrees of freedom
1. 2D Frames: NDOF = 3NJ – NR NR 3
2. 3D Frames: NDOF = 6NJ – NR NR 6
3. 2D Trusses: NDOF= 2NJ – NR NR 3
4. 3D Trusses: NDOF = 3NJ – NR NR 6
Where, NDOF is the number of degrees of freedom
In 2D analysis of frames some times axial deformation is ignored. Then NAC=No. of
axial condition is deducted from NDOF
Truss NM=11; NJ=6; NR =4; IND=NM+NR-2NJ IND= 11 + 4– 2 x 6 = 3 INDETERMINATE
Truss
NM=14; NJ=9; NR =4;
IND=NM+NR-2NJ
IND= 14+ 4 – 2 x 9 = 0
DETERMINATE
k)
l)
Fig 1.10
56
Examples:
1.2 Determine Degrees of Kinermatic Indeterminacy of the structures given below
Extensible
NJ=2; NR =3;
NDOF=3NJ-NR
NDOF=3 x 2 – 3= 3 (1, 2, 2)
Inextensible
NJ=2; NR =3; NAC=1
NDOF=3NJ-NR-NAC
NDOF=3 x 2 – 3-1= 2 (1, 2)
Extensible
NJ=4; NR =5;
NDOF=3NJ-NR
NDOF=3 x 4 – 5= 7
(1, 21, 23 3 ,y2,e1,e2)
Inextensible
NJ=4; NR =5; NAC=2
NDOF=3NJ-NR-NAC
NDOF=3 x 4 – 5-2= 5
(1, 21, 23 , 3 y2)
a)
b)
57
c)
Extensible
NJ=4; NR =6;
NDOF=3NJ-NR
NDOF=3 x 4 – 6= 6
(2, 3 , , e1,e2, e3)
Inextensible
NJ=4; NR =6; NAC=3
NDOF=3NJ-NR-NAC
NDOF=3 x 4 – 6-3= 3
(2, 3, )
58
d)
e)
f)
Extensible
NJ=4; NR =5;
NDOF=3NJ-NR
NDOF=3 x 4 – 5+1= 8
(21, 23, 3 , 2, 3, e1,e2, e3)
Inextensible
NJ=4; NR =5; NAC=3
NDOF=3NJ-NR-NAC
NDOF=3 x 4 – 5-3= 4
(21, 23, 3 , 2=3=)
Extensible
NJ=5; NR =6;
NDOF=3NJ-NR
NDOF=3 x 5 – 6 = 9
(2, 3, 4 , x2, x3, y3, x4 e1, e4)
Inextensible
NJ=5; NR =6; NAC=3
NDOF=3NJ-NR-NAC
NDOF=3 x 5 – 6 - 3= 6
(2, 3, 4 , x2= x4=1, x3, y3)
59
g)
1.5 Linear and Non Linear Structures
Structural frameworks are commonly made of wood, concrete or steel. Each of them has
different material properties that must be accounted for in the analysis and design. The
modulus of elasticity E of each material must be known for any displacement
computation. Typical stress-strain curve for these materials is shown in Fig.1.11. The
structure in which the stresses developed is within the elastic limit, and then the structure
is called Linear Structure. If the stress developed is in the plastic region, then the
NJ=6; NR =3; NDOF=2NJ-NR NDOF=2 x 6 – 3 = 9
A Truss
NJ=6; NR =4;
NDOF=2NJ-NR
NDOF=2 x 6 – 4 = 8
A Truss
60
structure is said to Non-Linear Structure. In addition to material nonlinearities, some
structures may behave in a nonlinear fashion due to change in the shape of the overall
structure. This requires that the structure displace an amount significant enough to affect
the equilibrium relations for the structure. When this occurs the structure is said to be
Geometrically nonlinear. Cable structures are susceptible to this type of nonlinearity. A
cantilever structure shown in Fig. 1.2 has geometrical nonlinearity
Fig.1.11 Stress-Strain Graph
61
Fig.1.12 Geometric Nonlinearity
62
Exercise Problems
P1.1 Determine Degrees of indeterminacy and classify the structures
P1.2 Determine Degrees of Kinematic indeterminacy
Reference Books
63
1. C.S.Reddy, Basic Structural Analysis,
Tata Mc Graw Hill, 2nd Edition, New Delhi, 1996.
2. Ashok.K.Jain, Elementary Structural Analysis, Newchand and brothers, Roorkee,
India, 1990
3. L.S.Negi and R.S.Jangid, Structural Analysis,
Tata Mc Graw Hill, 2nd Edition, New Delhi, 1997.
4. G.S.Pandit, S.P.Gupta and R.Gupta, Theory of Structures, Vol-1, Tata Mc Graw Hill,
2nd Edition, New Delhi, 1999.
5. Jeffrey P. Laible, Structural Analysis, Holt-Saunders International Editions, 1985
THREE HINGED ARCHES
An arch is a curved beam in which horizontal movement at the support is wholly or
partially prevented. Hence there will be horizontal thrust induced at the supports. The shape
of an arch doesn‟t change with loading and therefore some bending may occur.
Types of Arches On the basis of material used arches may be classified into and steel arches,
reinforced concrete arches, masonry arches etc.,
On the basis of structural behavior arches are classified as :
Three hinged arches:- Hinged at the supports and the crown.
Hinged at the
support
Hinged at the
crown
Springing
Rise
Span
Types of Arch
Material used
Structural behavior Loaded area
Loaded area
64
Two hinged arches:- Hinged only at the support.
Fixed arches:- The supports are fixed.
A 3-hinged arch is a statically determinate structure. A 2-hinged arch is an
indeterminate structure of degree of indeterminancy equal to 1. A fixed arch is a statically
indeterminate structure. The degree of indeterminancy is 3.
Depending upon the type of space between the loaded area and the rib arches can be
classified as open arch or closed arch (solid arch).
Hinges at the
support
Rib of the arch Rise
Span
Filled
Spandrel Open Arch
Braced
Spandrel Solid Arch
65
Comparison between an arch and a beam
axwL
wbxM
Arch -H x y
L
)ax(wwbxM
Beam
HyMMBeamArch
momentBeamArch HMM
Owing to its geometrical shape and proper supports, an arch supports loading with
less bending moment than a corresponding straight beam. However in case of arches there
will be horizontal reactions and axial thrust.
Analysis of 3-hinged arches It is the process of determining external reactions at the support and internal quantities
such as normal thrust, shear and bending moment at any section in the arch.
Procedure to find reactions at the supports Step 1. Sketch the arch with the loads and reactions at the support.
Apply equilibrium conditions namely 0,Fx 0Fy 0Mand
Apply the condition that BM about the hinge at the crown is zero (Moment of all the forces
either to the left or to the right of the crown).
Solve for unknown quantities.
W
E
C
E
B
E HB = H A
E HA = H
y RISE = h
SPAN
b a
VA = L
Wb x
VB = L
Wa
VA = L
Wb x
VB = L
Wa
H1 H1
W a b
x
66
P1: A 3-hinged arch has a span of 30m and a rise of 10m. The arch carries UDL of 0.6 kN/m
on the left half of the span. It also carries 2 concentrated loads of 1.6 kN and 1 kN at 5 m and
10 m from the „rt‟ end. Determine the reactions at the support. (sketch not given).
0Fx
0HHBA
BA
HH ------ (1)
To find vertical reaction.
0Fy
6.11
6.1115x6.0VV BA
------ (2)
0MA
kN35.7A
6.1125.4V
kN25.4V
05.7)15x6.0(20x125x6.130xV
A
A
B
B
B HB = 4.275
0.6 kN/m C
5 m 5 m
h = 10m
L = 30m VA = 7.35 VB = 4.25
HB = 4.275 A
1 kN 1.6 kN
67
To find horizontal reaction.
0MC
kN275.4H
kN275.4H
010xH15x25.410x6.15x1
A
B
B
OR
0MC
kN275.4H
kN275.4H
5.7)15x6.0(10xH15x375.7
B
A
A
To find total reaction
83.44H
Vtan
kN02.6VHR
82.59H
Vtan
kN5.8
35.7275.4
VHR
B
B1
B
2
B
2
BB
0
A
A1
A
22
2
A
2
AA
RA
A
A HA = 4.275 kN
VA = 7.35 kN RB
A HB = 4.275 kN
VB = 4.25 kN
68
2) A 3-hinged parabolic arch of span 50m and rise 15m carries a load of 10kN at quarter span
as shown in figure. Calculate total reaction at the hinges.
0Fx
BA
HH
To find vertical reaction.
0Fy
10VV BA ------ (1)
0MA
kN5.7VkN5.2V
05.12x1050xV
AB
B
To find Horizontal reaction
0MC
015H25VBB
To find total reaction.
RA
A
A HA = 4.17
VA = 7.5
RB
HB = 4.17
VB = 4.25
B HB
10 kN C
15 m
50 m VA VB
HA A
12.5 m
69
94.30H
Vtan
kN861.4R
VHR
92.60H
Vtan
kN581.8R
5.717.4R
HkN17.4H
0
B
B1
B
B
2
B
2
AB
0
A
A1
A
A
22
A
AB
3) Determine the reaction components at supports A and B for 3-hinged arch shown in fig.
To find Horizontal reaction
0Fx
0HHBA
BA
HH ------ (1)
To find vertical reaction.
0Fy
B HB
10 kN/m C
2.5 m
10 m
VA
VB
HA A
2 .4 m
180 kN
8 m 6 m
70
280VV
10x10180VV
BA
BA
------ (2)
0MA
33.1558V10H
3740V24H4.2
05x10x1018x1804.2xH24xV
BB
BB
BB
------ (3)
0MC
87.293V857.2H
144014V9.4xH
09.4xH14xV8x180
BB
BB
BB
------ (4)
Adding 2 and 3
AB
B
A
B
BB
HkN67.211H
33.1558177x10H
kN103V
kN177V
87.29333.1558V857.2V10
71
Bending moment diagram for a 3-hinged arch
We know that for an arch, bending moment at any point is equal to beam BM-Hy
(Refer comparison between arch and beam). Hy is called H-Moment. It varies with respect
to Y. Therefore the shape of BM due to Hy should be the shape of the arch. Therefore to
draw the BMD for an arch, draw the BMD for the beam over that superimpose the H-moment
diagram as shown in fig.
B
W C
h
a b
A
L
L
Wab
+
–
+ –
Shape – same
as arch BEAM B.M.
L
Wab
H – Moment diagram
Hh
Hh
72
Normal thrust and radial shear in an arch
Total force acting along the normal is called normal thrust and total force acting along
the radial direction is called radial shear. For the case shown in fig normal thrust
= + HA Cos + VA Cos (90 - )
= HA Cos + VA Sin
(Treat the force as +ve if it is acting towards the arch and -ve if it is away from the arch).
Radial shear = + HA Sin -VA Sin (90 - )
= HA Sin + VA Cos
(Treat force up the radial direction +ve and down the radial direction as -ve).
Note: 1) To determine normal trust and tangential shear at any point cut the arch into 2 parts.
Consider any 1 part. Determine net horizontal and vertical force on to the section.
Using these forces calculate normal thrust and tangential shear.
2. Parabolic arch: If the shape of the arch is parabolic then it is called parabolic arch.
If A is the origin then the equation of the parabola is given by y = cx [L – x] where C
is a constant.
We have at X = 2
L y = h
B
C
h
A Horizontal
y
x 2
Normal
A
Radial
Normal
VERTICAL
NORMAL
HORIZONT
AL
VA
HA
M
A HA
VA RADIAL
Normal Radial
90 -
73
h = C 2
L.
2
L.C
2
LL
2
L
2L
H4C
Equation of parabola is
xLL
hx4y
2
is given by the following equation.
2
2xLx
L
h4y
x2LL
h4
dx
dytan
2
x2LL
h4tan
2
1. A UDL of 4kN/m covers left half span of 3-hinged parabolic arch of span 36m and central
rise 8m. Determine the horizontal thrust also find (i) BM (ii) Shear force (iii) Normal
thrust (iv) Radial shear at the loaded quarter point. Sketch BMD.
0Fx
BA
BA
HH
0HH
------ (1)
0Fy
72VV
18x4VV
BA
BA
------ (2)
0MA
4 kN/m C
h = 8 m
36 m VA VB
HA A B HB
74
kN54VkN18V
09x18x436xV
AB
B
0Mc
kN5.40H
kN5.40H
08xH18xV
A
B
BB
BM at M =
- 40.5 x 6 + 54 x 9 xLL
hx4y
2
- 4 x 9 x 4.5 93636
9x8x4y
2
= 81 kN.m m6y
Shear force at M = + 54 – 4 x 9 = 18 kN (only vertical forces)
tan x2LL
h42
= 9x23636
8x42
96.230
Normal thrust = N = + 40.5 Cos 23.96 + 18 Cos 66.04
= 44.32 kN
S = 40.5 Sin 23.96 – 18 Sin 66.04
S = - 0.0019 0
VERTICAL
NORMAL
HORIZONTAL
66.04
40.5 kN
= 23.96 M
A
40.5 kN RADIAL
4 kN/m
54 kN
9 m
y = 6
18 kN
75
m5.13x
x354x
18
x18
54
x
54 x 13.5 – 4 x 13.5 2
5.13x
kNm5.364
A symmetrical 3-hinged parabolic arch has a span of 20m. It carries UDL of intensity 10
kNm over the entire span and 2 point loads of 40 kN each at 2m and 5m from left support.
Compute the reactions. Also find BM, radial shear and normal thrust at a section 4m from
left end take central rise as 4m.
+ –
364.5 kNm 324 kNm
A C B 13.5 m
13.5
A B
36 m 54 kN 18 kN
18 – x
x = 13.5
18 kN 18 kN
324 kN/m
Beam BM
364.5 kN/m kN/m
+
4 kN/m
54 kN
76
0Fx
BA
BA
HH
0HH
------ (1)
0Fy
280VV
020x104040VV
BA
BA
------ (2)
0MA
kN166V
kN114V
020xV10)20x10(5x402x40
A
B
B
0Mc
C
4 m
20 m
40 kN
M
2 m 3m
10 kN/m 40 kN
77
kN160H
kN160H
010x1144xH5)10x10(
A
B
B
BM at M
= - 160 x 2.56
+ 166 x 4 – 40 x 2
- (10 x 4)2
= + 94.4 kNm
xLL
hx4y
2
42020
4x4x42
m56.2y
tan x2LL
h42
= 4x22020
4x42
64.250
Normal thrust = N = + 160 Cos 25.64
+ 86 Cos 64.36
= 181.46 kN
S = 160 Sin 25.64
- 86 x Sin 64.36
S = - 8.29 kN
VERTICAL
NORMAL
HORIZONT
AL
64.35
160 kN
= 25.64 M
REDIAL
10 kN/m
160 kN
4 m
y = 2.56
86 kN
40 kN
2 m
166 kN
78
Segmental arch
A segmental arch is a part of circular curve. For such arches y =2L
)xL(hx4 is not
applicable since the equation is applicable only for parabolic arches. Similarly equation for
will be different.
To develop necessary equations for 3-hinged segmental arch
Relationship between R, L and h:
From OAD
222 ODADOA
2
2
22
2
2
2
2
2
h4
LRh2
hRh2R4
LR
hR2
LR
2
h
h8
LR
2
R
xSin
OM
OECos
R
yhRCos
B
C
h (90-)x
A
O
R R
Origin
E M
90
x
y
L/2 D L/2
ROBOCOA
ROM
79
1) A 3-hinged segmental arch has a span of 50m and a rise of 8m. A 100 kN load is acting at
a point 15 m from the right support. Find horizontal thrust at the supports (ii) BM,
Normal thrust and radial shear at a section 15 m from the left support.
Reactions:
0Fx
BS HH ------ (1)
0Fy
kN100VV BA ------ (2)
0MA
kN30V
kN70V
035x10050xV
A
B
B
0Mc
BA
AA
HkN75.93H
08xH25xV
NORMAL 30 kN
76.97
93.75 kN
= 13.43
93.75 kN REDIAL
30 kN
15 m
y
Z
B HB 93.75 kN
C
Origin
25 m
VA = 30 kN VB = 70 kN
HA = 93.75 kN A
100 kN
15 m Z
25 m
15 m
80
m06.43R
2
8
8x8
50
2
h
h8
LR
2
2
43.13
06.43
10
R
x Sin
* In case of segmental arch A is not origin
R
yhRCos
m822.6y
06.43
y806.43 13.43 Cos
B.Mz = 30 x 15 – 93.75 x 6.822
= - 189.562 kNm
N = 93.75 Cos 13.43 + 30 Cos 76.57
N = 98.15 kN
S = 93.75 Sin 13.43 – 30 Sin 76.57
= - 7.41 kN
CABLES AND SUSPENSION BRIDGES Cables are used to support loads over long spans such as suspension bridges, roof of
large open buildings etc. The only force in the cable is direct tension. Since the cables are
flexible they carry zero B.M.
Analysis of cables Analysis of cable involves determination of reactions at the support and tension over
different parts of the cable.
To determine the reactions at the support and tension equilibrium conditions are used.
In addition to that BM about any point of the cable can be equated to zero.
P1: Determine the reactions components and tension in different parts of the cable shown in
figure. Also find the sag at D and E.
81
0Fx
BA
BA
HH
0HH
------ (1)
0Fy
kN120VV BA ------ (2)
0MA
kN55V
kN65V
080xV60x5040x4020x30
A
B
B
0Mc (About the point where sag is given)
BA
A
HkN110H
020x5510xH
Point A:
0Fx
A 20 m 20 m 20 m 20 m B
1
T1 10 m yD = 14.55
yE = 11.82
HA = 220 kN HB = 110 kN
VA = 55 kN VB = 65 kN T4
T2
T3
30 kN 40 kN
50 kN
C
D
E 2
= 26.56
55 kN
110 kN
T1
82
kN123T
011056.26CosT
1
1
tan 20
101
56.261
To find YE:-
We have 0M.B E
m82.11Y
020x65Yx110
E
E
To Find YD:-
0M.BD
m55.14Y
0YxH40x6520x50
D
DB
83
Point D:
0Fx
984.0TT
081.12CosT77.7CosT
23
23
tan 20
1055.142
81.121
tan20
82.1155.143
77.73
0FY
kN95.110T
kN75.112T
04081.12SinT77.7Sin984.0T
040
81.12SinT77.7SinT
3
2
22
21
12.81 =
40 kN
110 kN
3 = 7.77
T2 T3
84
Part B:
0Fx
kN77.127T
058.30CosT110
4
4
tan 20
82.114
58.304
Note: It can be observed that more the inclination or the slope more is the tension. Near the
supports slope will be more and hence tension will also be more.
2) A Chord Supported at its ends 40 m apart carries loads of 200 kN, 100 kN and 120 kN at
distances 10 m, 20 m and 30 m from the left end. If the point on the chord where 100 kN
load is supported is 13 m below the level of the end supports. Determine
(a) Reactions at the support.
(b) Tension in different part.
(c) Length of the chord.
0Fx
A 10 m 10 m 10 m 10 m E
T1 yB 13 m
yD
HA = 200 HB = 200
VA = 230 VE = 190 T4
T2 T3
200 kN 100 kN
120 kN
B
C
D
1
2
3
3
110 kN
65 kN
30.18
T4
85
EA
EA
HH
0HH
------ (1)
0Fy
kN420VVEA ------ (2)
0MA
kN230V
kN190V
040xV30x12020x10010x200
A
E
E
0Mc (Always take BM about the point for which sag is given)
EA
A
HkN200H
020x23010x20013xH
86
Point A:
0M.B B
m5.11Y
0Yx20010x230
B
B
49
10
5.11tan
1
1
1
0Fx
kN85.304T
020049CosT
1
1
Point B:
0Fx
2
200 kN
204.85 kN
T2
49°
49°
230 kN
200 kN
T1
87
kN22.202T
049Cos
8.30453.8CosT
2
2
tan 2 = 10
11.5 13
2 = 8.53
Point C:
0M.BD
m5.9Y
010x190Yx200
D
D
0Fx
kN90.211T
053.8Cos22.2023.19CosT
3
3
tan
10
5.9133
3.193
100 kN
202.22 kN
T3
8.53 19.3
88
Point E:
0Fx
kN85.275T
053.43CosT200
4
4
tan
10
5.94
53.434
Length of the Chord = DECDBCAB
Cos AB
101
49Cos
10AB
m22.15AB
Cos BC
102
53.8Cos
10BC
m12.10BC
CosCD
103
3.19Cos
10CD
m6.10CD
Cos DE
104
200 kN
190 kN
4
T4
89
53.43Cos
10DE
m8.13DE
Total length of Chord = 49.76m
3) Determine reactions at supports and tension indifferent parts of the cable shown in
figure.
0Fx
BA
BA
HH
0HH
------ (1)
0Fy
kN120VVBA ------ (2)
0MA
6250V100H15
015xH100xV75x4050x5025x30
BB
BB
67.416V67.6HBB ------ (3)
0M.B D
1000V50H5.2
025x4050xV5.2xH
BB
BB
400V20HBB
------ (4)
HA A 25 m 25 m 25 m 25 m 1
3
2
T4 T3
T2
T1
VA
VB
40 kN
50 kN
30 kN
C
D
E
8.75
7.5 m determined
10 m
(given)
8.95
26.25
15 m
2.5
90
(3) – (4) gives
40067.416V67.26B
kN62.30VB
ABHkN42.212H
kN38.89VA
Point A:
0Fx
kN05.225T
042.21229.19CosT
1
1
25
75.8tan 1
1
29.191
1 = 19.29°
89.38 kN
212.42 kN
T1
91
30 kN
225.05 kN
T2
19.29
2 = 5.71
Point C:
25
5.2tan 1
2
71.52
25
75.8tan 1
3
29.193
25
25.11tan 1
4
22.244
4) A cable is used to support loads 40 kN and 60 kN across a span of 45 m as shown in
figure. The length of the cable is 46.5 m. Determine tension in various segments.
0Fx
DA
DA
HH
0HH
------ (1)
0Fy
kN100VVDA ------ (2)
0MA
kN67.46V
kN33.53V
045xV30x6015x40
A
D
D
------ (3)
0M.B B
05.700YH
0YxH15x67.46
BA
BA
A 15 m 15 m 15 m D
T1
yB = 4.4 m yC = 5.025
HA = 159.2 kN HD
VA = 46.67 = 0 VD = 53.33
T2
T3
40 kN
60 kN
B
D
92
0MC
10.800YH
0YxH15x4030x67.46
CA
CA
10.800
05.700
YH
YH
CA
BA
CBY875.0Y
We have DECDBCAB = 46.5m
m5.46Y15)YY(15Y152
C
22
BC
22
B
2
21
21 /2
CC
2/2
C
2 Y875.0Y15Y875.015
5.46Y15
Y0156.015Y766.015
5.46Y15
21
21
21
21
/2
C
2
/2
C
2/2
C
2
/2
C
2
5.4615
Y1
15
Y0156.01
15
Y766.0115
21
21
21 /
2
2
C
/
2
2
C
/
2
2
C
5.4615
Y
2
11
15
Y0156.0x
2
11
15
Y766.0x
2
1115
2
2
C
2
2
C
2
2
C
15
5.46Y00396.03
2
C
1.0Y00396.02
C
m4.4Y
m025.5Y
B
C
We have
DA
A
BA
HkN2.159H
4.4
05.700H
05.700YH
15
4.4tan 1
1
35.160
1
93
Point A:
0Fx
kN90.5.16T
0kN2.159
35.16CosT
1
0
1
Point C:
0Fx
23
23
T053.1T
0385.2CosT52.18CosT
15
4.4025.5tan 1
2
385.22
15
025.5tan 1
3
52.183
1
46.67 kN
159.2 kN
T1
94
0FY
kN89.167T
kN44.159T
60T376.0
060385.2SinT52.18SinT053.1
060385.2SinT52.18SinT
3
2
2
22
23
General Equation of a cable or Differential Equation.
General shape of a cable depends on nature of loading, location of loads, type of
supports etc. The equilibrium of a part of a cable shall be considered to obtain equation for
cable when the cable is subjected to all over UDL.
Let us consider the equilibrium of a small length „ds‟ of the cable shown in figure.
Let the cable be subjected to UDL of intensity W over horizontal span figure shows tension
and horizontal, vertical reactions in the part of the cable we have.
60 kN
T2 T3
2 3
95
H = T Cos
V = T Sin
SecHCos
HT
tan
CosT
SinT
H
V
V = H tan
Let us consider the equilibrium of the part of the cable shown in fig.
We have.
0FY
(V + dv) – V – W x dx = 0
dv = wdx
W
dx
tanHd
Wdx
dx
dyHd
H
W
dx
Yd2
2
To derive equations for cable profile and tension in the cable when it is supported at
the same level and subjected to horizontal UDL.
Let us consider a cable of span L and max sag H subjected to UDL of intensity „W‟ as
shown in fig. From general equation we have H
W
dx
d
2
2
y
C
D
W kN/m
VD = v + dv T + d
HD = H
V = VC = TSin T
C
H = HC = TCos
ds
+ 2
dy
dx
96
Integrating with respect to x we have 1
ycx.
H
W
dx
d
We have at X = 0 0dx
dy
Substituting 0 = .H
W0 + C1
C1 = 0
x.H
W
dx
dy
Integrating with respect to x we get 2
2
C2
X.
H
WY
We have at X=0 Y=0
2
C00
2
C =0
2
X.
H
WY
2
At X= 2/ Y = h
Substituting
2
2/.
H
Wh
2
h8
WH
2
We have 2
X.
h8
w
WY
2
2
2
2x
L
h4Y
If A is considered as the origin then
)xL(L
hx4Y
2
W
y L A B
h
x
97
To fin the tension at any point on the cable:
Tension at any point
T = H Sec
= H 2Sec
= H 2tan1
T = H
2
dx
dy1
To derive an expression for cable profile when it is subjected to horizontal UDL and
supports are at different levels:
General equation for cable profile is
H
W
dx
d
2
2
y
Let us consider each part separately we have
H2
WxY
2
At x = -L1, Y = b
H2
WLbOR
H2
LWb
2
1
2
1
------ (1)
At X = L2, Y= a + b
H2
WLba
2
2 ------ (2)
2
2
2
1
L
L
ba
b
ba
b
L
L
2
1
Add (1) on either side
w
a
L1
B
b
L2 y
x
L1
O
98
1ba
b1
L
L
2
1
)LLL(
ba
bab
L
LL21
2
21
ba
bab
L
L
2
bab
baLL
2
Substituting in (2)
22
)ba(b
)Ba(L
H2
W b)(a
22
bab2
WLH
Tension at any point is given by
T = H Sec
T = H 2Sec
T = H 2tan1
= H
2
dx
dy1
T = H
2
H
wx1
Y = H2
wx 2
H2
wx2
dx
dy
To derive an expression for length of the parabolic cable profile when the supports are
at the same level
99
We have ds =22 dydx
Ds = dx
2
dy
dx1
When the supports are at the same level we have
2
2
L
hx4Y
2L
hx8
dx
dy (taking origin at the center)
ds = dx 4
22
L
xh641
Total length of the cable is given by
LC = dxL
xh641ds
½
4
222/
2/
= 2 dxL
xh641
½
4
222/
0
= 2 dx............L
xh64
2
11
4
222/
0
LC = 2
2/
0
3
4
2
3
x
L
h32x
(expanding by binomial theorem)
LC = 2
24
L
L
h32
2
L 3
4
2
LC = L3
h8L
2
To derive an expression for length of the cable profile when the supports are at
different levels
dx
dy
ds
100
LC = Length of cable
Between A & C + Length of cable between C&B
LC = 2
1 Length of cable of span 2L1 +
2
1 length of cable of span 2L2
But LC = L +
2
2
2
1
2
1C
2
L2
ba.
3
8L22/1
L2
b
3
8L2
2
1Le.i
L3
h8
2
2
2
1
2
1L
)ba(
3
2L
L
b
3
2L
LC =
2
2
1
2
21L
)ba(
L
b
3
2LL
1) A cable suspends across a gap of 250m and carries UDL of 5kN/m horizontally calculate
the maximum tensions if the maximum sag is 1/25
th of the span. Also calculate the sag at
50m from left end.
We have
T = H
2
dx
dy1
Take origin as left end
xLL
hx4Y
2
x2LL
h4
dx
dy2
Tension is maximum at the supports i.e., at X = 0 from A
5 kN/m
250 m A B
h = 250x25
1
= 10 m
a L1
B
b
L2
c
A
101
02250250
104
dx
dy2
16.0dx
dy
kN25.3906108
2505
h8
WLH
22
kN93.395516.0125.3906T 2
max
We have )xL(L
hx4Y
2 [x measured from A]
)50250(250
50104Y
2
m4.6Y
2) Determine the length of the cable and max tension developed if the cable supports a load
of 2kN/m on a horizontal span of 300m. The maximum sag is 25m
h8
WLH
2
258
3002 2
kN900H
2
dx
dy1HT
xLL
hx4Y
2 [X from A]
x2LL
h4
dx
dyY
2
Tension is maximum at the supports i.e., at x = 0
02300300
254
dx
dy2
2 kN/m
300 m A B
25
102
333.0dx
dy
2
max333.01900T
kN68.948Tmax
Length of cable LC = L3
h8L
2
3003
258300
2
m55.305LC
3) Determine the maximum span for a mild steel cable between supports at the same level if
the central dip. is 1/10th of the span and permissible stress in steel is 150 N/mm
2. Steel
weighs 78.6 kN/m3. Assume the cable to hang in a parabola.
Here the weight of the cable itself is acting as UDL on the cable. We have SP weight
=Volume
Wieght
Weight = Specific Weight x Volume
= Specific Weight x Area x Length
Length
Weight = Specific Weight x Area
m
kNA6.78
mAm
kN6.78W 2
3
We have h8
WLH
2
L10
18
LA6.78 2
LA25.98H
2
dx
dy1HT
)XL(L
hx4Y
2
L
A
103
)x2L(L
h4
dx
dy2
Tension is maximum at X = 0
L
h4
dx
dy
2
maxL
L10
14
1AL25.98T
AL81.105Tmax
L81.105A
Tmax
L81.105Fmax
2
3
2max
m
kN10150
MPa150
mm
N150F
m63.1417L
L81.10510150 3
Bridges supported by cables
Suspension Bridge
Anchor
cable
104
Anchoring of cables There are 2 methods by which suspension cable can be anchored:
1) Continuous cable or pulley type anchoring.
2) Non- Continuous cable or saddle type anchoring.
Continuous cable or pulley type anchoring In this method suspension cable itself passes over roller or guide pulley on the top of
the tower or abutment and then anchored. The tension remains same in the suspension cable
and anchor cable at the supports.
2
A
2
AA VHT
A
A1
H
Vtan
ht
Tower
B
TA
TA
Fan Type Cable stayed Bridge
Harp type Cable stayed Bridge
Pulley
Anchor cable
Suspension cable
Abutment
A
TA
TA
105
is the inclination of the suspension cable with the horizontal. Net horizontal force
on tower HT = TA Cos ~ TA Cos
Where ht is the height of the tower.
2) Saddle type anchoring or Non-continuous cable In this method of anchoring suspension cable are attached to saddles mounted on
rollers on the top of the tower as result in suspension cable and anchor cable will be differed.
However horizontal components of tension will be equal.
B
TA
Ta
Anchor cable
Suspension cable
Abutment
106
2
A
2
AA VHT
A
A1
H
Vtan
0M
hHM
SinTSinTV
HCosTCosT
tT
aAT
TaA
1) A cable of span 150m and dip 15m carries a load of 6 kN/m on horizontal span. Find the
maximum tension for the cable at the supports. Find the forces transmitted to the supported
pier if
(a) Cable is passed over smooth rollers or pulleys over the pier.
(b) Cable is clamped to saddle with smooth rollers resting on the top of the pier.
For each of the above case anchor cable is 30 to horizontal. If the supporting pier is
20m tall. Determine the maximum BM on the pier.
kN1125
158
1506
h8
WHH
2
2
BA
kN4502
1506VV BA
8.21
1125
450tan
H
Vtan
kN66.1211VHTT
0
1
A
A1
2
A
2
AAmax
6 kN/m
150 m A B
15 m VB
HB HA
VA
107
Case 1: Cable over smooth pulley
HT = 1211.66 Cos 21.8
1211.66 Cos 300
= 75.73kN
VT = 1211.66 Sin 21.8 + 1211.66 Sin 300
VT = 1055.77kN
M = HT X ht
= 75.73 X 20 = 1513.4kNm
Case 2 : Cable clamped to saddle
Here TA Cos CosTa
1211.66 Cos 21.8 = Ta Cos 30
Ta 1299.05kN
In this case HT = 0
M =
030Sin05.12998.21Sin66.1211
SinTSinTV aAT
VT = 1099.5 kN
3) A Suspension cable is suspended from 2 pier A and B 200m apart, B being 5m below
A the cable carries UDL of 20kN/m and its lowest point is 10m below B. The ends of
the cable are attached to saddles on rollers at the top of the piers and backstays anchor
cables. Backstays may be assumed to be straight and inclined at 600 to vertical.
Determine maximum tension in the cable, tension in backstay and thrust on each pier.
Let C be the origin
Tower
300 =
TA = 1211.66 kN
TA
= 21.8
y
5 m
C
200 m
10 m
20 kN/m
VA
= 2202.05
AA
= 8081.6 kN
L1 L2
= 110.12 m = 89.89 m
x
300
TA = 1211.66 kN
Ta
21.8
108
We have H2
WxY
2
m10YLxAt
m15YLxAt
2
1
H2
)L(w15
2
1
H2
)L(w10
2
2
2
2
12
L
L
10
15
21
2
1
L225.1L
10
15
L
L
But 200 L2L1
m12.110L
m89.89L
200LL225.1
1
2
22
0Fx
BA
BA
HH
0HH
---- (1)
0Fy
4000VVBA ---- (2)
0MA
80000V40H
400000V200H5
05H200V10020020
BB
BB
BB
---- (3)
300
TA = 8376.26 kN
Ta
15.24
600
109
0MC
212.88080V989.48
212.8080V989.8H
12.80802V89.89H10
010H89.89V2
89.89)89.8920(
B
BB
BB
BB
----- (4)
kN05.2202V
kN6.8081H
kN6.8081H
kN95.1797V
A
A
B
B
Since VA> VB tension at A is maximum
kN94.9331T
30CosT24.15Cos26.8376
CosTCosT
24.15
6.8081
05.2202tan
H
Vtan
kN23.8376T
6.808105.2202
VHTT
a
0
a
aA
1
A
A1
A
22
A2
A2
Amax
110
kN77.6867V
30Sin94.933124.15Sin26.8376
SinTSinTV
T
aAT
ANALYSIS OF CONTINUOUS BEAMS (using 3-moment equation)
Stability of structure
If the equilibrium and geometry of structure is maintained under the action of forces
than the structure is said to be stable.
External stability of the structure is provided by the reaction at the supports. Internal
stability is provided by proper design and geometry of the member of the structure.
Statically determinate and indeterminate structures
A structure whose reactions at the support can be determined using available
condition of equilibrium is called statically determinate otherwise it is called statically
indeterminate.
Ex:
W
A B
HA HB
VA VB
MA MB
End moments
FIXED BEAM
111
No. of unknowns = 6
No. of eq . Condition = 3
Therefore statically indeterminate
Degree of indeterminacy =6 – 3 = 3
No. of unknowns = 3
No. of equilibrium Conditions = 2
Therefore Statically indeterminate
Degree of indeterminacy = 1
Advantages of Fixed Ends or Fixed Supports
1. Slope at the ends is zero.
2. Fixed beams are stiffer, stronger and more stable than SSB.
3. In case of fixed beams, fixed end moments will reduce the BM in each section.
4. The maximum defection is reduced.
Bending Moment Diagram for Fixed Beam
Draw free BMD
Draw fixed end moment diagram, superimpose one above the other.
W W
A
RA
RB
A C
RC
112
Ex:
Continuous beams
Beams placed on more than 2 supports are called continuous beams. Continuous
beams are used when the span of the beam is very large, deflection under each rigid support
will be equal zero.
BMD for Continuous beams
BMD for continuous beams can be obtained by superimposing the fixed end moments
diagram over the free bending moment diagram.
Three - moment Equation for continuous beams OR CLAYPREON‟S
THREE MOMENT EQUATION
Ex:
W
4
WL
2
L
2
L
+
+
M M
113
FREE B.M.
1x
a1 a2
8
2WL
L2 L2
A B C
N N
4
WL
2x
114
22
2
C
22
2
11
1
B
11
1
AIE
LM
IE
L
IE
LM2
IE
LM
2
BC
1
BA
222
22
111
11
LL6
LIE
xa6
LIE
xa6
The above equation is called generalized 3-moments Equation.
MA, MB and MC are support moments E1, E2 Young‟s modulus of Elasticity of 2 spans.
I1, I2 M O I of 2 spans,
a1, a2 Areas of free B.M.D.
21 xandx Distance of free B.M.D. from the end supports, or outer supports.
(A and C)
A, B and C are sinking or settlements of support from their initial position.
Normally Young‟s modulus of Elasticity will be same through out than the equation
reducers to
2
2C
2
2
1
1B
1
1A
I
LM
I
L
I
LM2
I
LM
2
BC
1
BA
22
22
11
11
LL6
LI
xa6
LI
xa6
If the supports are rigid then A = B = C = 0
2
2C
2
2
1
1B
1
1A
I
LM
I
L
I
LM2
I
LM
22
22
11
11
LI
xa6
LI
xa6
If the section is uniform through out
2C21B1A LMLLM2LM 2
22
1
11
L
xa6
L
xa6
115
Note:
1. If the end supports or simple supports then MA = MC = 0
2.
MC = - WL3
If three is overhang portion then support moment near the overhang can be computed
directly.
3.
If the end supports are fixed assume an extended span of zero length and apply 3-
moment equation.
Zero
Span
A
B
C
A1 D
Zero
Span
A B C
N N
D
L1 L2 L3
A B C
N N
116
Note: i)
In this case centroid lies as shown in figure
ii)
a b
Wa Wa
2
Lx
L
a b W
L
Wab
a b
3
a
3
b
117
Problems
1. For the continuous beam shown in fig. draw BMD and SFD. Assume
uniform cross section.
(Take care of ordinates)
A
B
C
20 kN 10 kN/N
RA = 6.25 RC = 11.25 RB = 32.5
1.5 m 1.5 m 3 m
1.5 m 1.5 m
FREE
BMD 11.25 kNm
+
–
– +
+
END MOMENT
DIAGRAM
15 kN 11.25 kNm
11.25 kNm
O O
15 kNm
11.25 kNm
118
To Calculate Bending Moment
Span AB
4
LW x =
4
3 x 20
= 15 kNm
Span BC
8
LW x 2
= 8
3 x 10 2
= 11.25 kNm
15x3x2
1a1
a1 = 22.5
5.11 x
25.11x3x3
2a 2
a2 = 22.5
5.1x2
MA = MC = 0
Applying 3 – moment equation.
2211 2 LMLLMLM CBA
2
22
1
11 66
L
xa
L
xa
3
5.15.226
3
5.15.226332
xxxxM B
13512 BM
kNmM B 25.11
5.1)310(3 xxRM CB
- 11.25 = 3RC - 45
RC = 11.25 kN
5.1x203xRM AB
- 11.25 = 3RA - 30
RA = 6.25 kN
119
Fy = 0
6RA + RB + RC = 20 + 30
6.25 + RB + 11.25 = 50
RB = 32.5 kN
A
B
C
20 kN 10 kN/N
RA = 6.25 kN RC = 11.25 kN RB = 32.5 kN
1.5 m 1.5 m 3 m
+ +
– –
6.25 6.25
13.75 13.75 11.25 kN
18.75 kN
120
2. Draw BMD and SFD for the continuous beam shown in Fig.
A
B
D
4 kN/N
RA = 18.98 kN RD = 18.63 RC = 48.58
8 m 10 m 6 m
kNm48
FREE BMD
32.19
+
–
END MOMENT DIAGRAM O O
6 kN/N 8 kN/N
C
RB = 49.81
kNm50 kNm36
40.16
+
48 kN 50 kNm 36 kNm 40.16
32.19 + +
+ +
– – –
20.79 kNm 29.37 kNm
29.02 18.63
18.98
121
To Calculate Bending Moment
Span AB
8
LW x 2
= 8
8 x 6 2
= 48 kNm
Span BC
8
LW x 2
= 8
10 x 4 2
= 50 kNm
Span CD
8
LW x 2
= 8
6 x 8 2
= 36 kNm
48x8x3
2a1
a1 = 256
41 x
50x10x3
2a 2
333.33
52 x
MA = MD = 0
Applying 3 – moment equation between A, B and C or for spans AB and BC.
2C21B1A LMLLM2LM
2
22
1
11
L
xa6
L
xa6
10
5x33.333x6
8
4x256x610M108M2 CB
99.17671036 CB MM
11.49277.0 CB MM ----- (1)
122
Applying 3 – moment equation between A, B and D.
2D21C1B LMLLM2LM 2
22
1
11
L
xa6
L
xa6
6
31446
10
533.3336610210
xxxxMxM CB
99.14313210 CB MM
199.1432.3 CB MM ----- (2)
Solving (1) and (2)
(1) and (2)
- 2.923 MC = - 94.09
MC = - 32.18 kNm
MB = 40.16 kNm
MC = RD x 6 – (8 x 6)3
- 32.2 = 6RD - 144
RD = 18.63 kN
MB – RA x 8 – (6 x 8) 4
- 40.16 = 8RA – 192
RA = 18.98 kN
MC = 18.98 x 18 – (6 x 8) (14) + RB x 10 – (4 x 10) 5
RB = 49.81 kN
Fy = 0
RA + RB + RC + RD = 48 + 40 + 48
RC = 48.58 kN
50x10x3
2a1
a1 = 333.33
mx 51
36x6x3
2a 2
a2 = 144
mx 32
123
3. Draw SFD and BMD for the continuous beam shown in Fig.
A B
RA = 10.73 kN RD = 18.63
2 m 5 m
FREE BMD –
C
RB = 23.7225 kN
kNm16
+
16 kNm
19.62 kNm 30 kNm
–
+
16 kN
kNm30
19.62 kNm
+ +
– –
12.4525
11.27 SFD 3.548 kN
0.73
3.548
12 kN
124
To Calculate Bending Moment
Span AB
L
Wab =
6
2 x4 x 12 = 16 kNm
Span BC
L
Wab =
8
5 x3 x 16 = 30 kNm
1662
11 xxa
a1 = 48
3
11
aLx
3
3461
x
3082
12 xxa
a2 = 120
3
bLx 2
2
3
582
x
mx 33.42
MA = 0 MC = 0
Applying 3 – moment equation between A, B and C.
2C21B1A LMLLM2LM
2
22
1
11
L
xa6
L
xa6
8
33.4x120x6
6
33.3x48x686M2 B
kNmM B 62.19
MB = - 16 x 3 + RC x 8
125
- 19.62 = - 48 + 8RC
RC = 3.5475 kN
MB = RA x 6 – 12 x 2
- 19.62 = 6RA – 24
RA = 0.73 kN
RA + RB + RC = 28
RB = 23.7225 kN
126
4. Draw BMD and SFD for the continuous beam shown in figure clearly indicate all salient
points.
Applying 3 – moment equation between A', A and C.
2C21A1'A LMLLM2LM 2
22
1
11 66
l
xa
l
xa
4
2x33.149x64xM)40(Mx2 CA
44848 CA MM ----- (1)
Applying 3 – moment equation between A, C and E.
A
C
E
16 kN 20 kN/N
RA = 46.235 RE = 6.33 RC = 83.435
2 m 2 m 2 m 4 m
kNm33.536
4x2x40
1.5 m 1.5 m
56 kNm
56 kNm
–
–
+
+
34.98
kNm
O
40 kN
B D
42.04 53.33 kNm
+ +
– –
6.235
9.765 kN 6.33
46.235
6.33
49.765 kN
33.67 33.67
A
Zero
56x4x3
2a 2
a2 = 149.33
mx 22
127
2E21C1A LMLLM2LxM 2
22
1
11 66
L
xa
L
xa
But ME = 0
MA x 4 + 2MC (4 + 6) + 0
6
33.3x160x6
4
2x33.149x6
4MA + 20MC = - 980.79 ------ (2)
x by 2
8MA + 40MC = - 1961.58 ------ (3)
(3) – (1)
58.151336 CM
kNmMC 04.42
kNmM A 99.34
MC = RA x 4 – 16 x 2 – 20 x 4 x 2 – 34.98
- 42.04 = 4RA – 32 – 160 – 34.98
RA = 46.235 kN
Fy = 0
RA + RC + RE = 16 + 40 + (20 x 4)
RC = 83.435 kN
MC = - 40 x 2 + RE x 6
- 42.04 = - 80 + 6RE
RE = 6.33 kN
A C
16 kN 20 kN/N
48 kN 48 kN
56 kNm
B
O O
MB = 48 x 2 – (20 x 2)
= 96 - 40
= 56 kNm
56x4x3
2a1
a1 = 149.33
mx 21
33.5662
12 xxa
= 1690
3
bLx 2
2
3
462
x
mx 33.32
128
5. Analyse the continuous beam shown in figure and draw BMD and SFD.
MB = RA x 4 – 60 x 3 -30.25
= - 18.24 x 4RA – 180 - 30.25
RA = 48 kN
MB = - 20 x 4 x 2 + RC x 4 - 30 x 5
RC = 72.94 kN
Fy = 0
RA + RB + RC = 30 + 60 + 20 x 4
RB = 49.06 kN
There is overhang portion CD
MC = - 30 x 1
+
A
C
E
60 kN 20 kN//m
RA = 48kN RC = 72.94 RB = 49.06
1 m 3 m 4 m 1 m
kNm454
3x1x60
40 kNm
40 kNm
–
+
18.24
B D
+ +
– –
48 kN
12 kN
48
42.94
37.06 30 kN
30 kN
Zero
3I 4I
CD
4I
–
+ 30.25
45
–
30 kNm
+ 30 kN
O
A'
+
hxbxa2
12
45x4x2
1
a2 = 90
3
bLx 2
2
3
34
= 2.333 m
129
MC = - 30 kNm
Applying 3 – moment equation between A',A and B.
2
2B
21
21
A
1
1A
I
LM
II
LLM2
I
LM
22
22
11
11 66
IL
xa
IL
xa
Ix
xx
IM
IM BA
34
333.2906
3
4
3
42
95.31448 BA MM
37.395.0 BA MM ----- (1)
Applying 3 – moment equation between A, B and C.
2
2
21
21
1
1 2I
LM
II
LLM
I
LMCB
A
22
22
11
11 66
IL
xa
IL
xa
IM
IIM
I
xMCB
A
4
4
43
442
3
4
Ix
xx
Ix
xx
44
267.1066
34
67.1906
1.333 MA + 4.667MB + MC
= - 75.15 – 80.00
MA + 3.509MB + 94.099 ----- (2)
Saving (1) and (2)
MB = - 18.24 kNm
MA = - 30.25 kNm
6. Draw SFD and BMD for the beam shown in figure.
45x4x2
1a1
= 90
3
11
aLx
3
14x1
= 1.67m
40x4x3
2a 2
a2 = 106.67
mx 22
130
Fy = 0
RB + RC = 0
MB = 0
120 – RC x 7 = 0
RC = 17.14 kN
RB = 17.14 kN
BM at C = 0
BM at B = 0
BM at E Just to left - 17.14 x 4 = - 68.56
BM at E Just to right - 17.14 x 3 = + 51.46 kNm
Applying 3 – moment equation between A, A and B.
A
C
6 kN/m
10 m 4 m 3 m
B
C'
120 kNm
ZERO ZERO
A'
75
51.46 67.85
68.56 kNm
O O
51.46 kNm75
8
10x6 2
+
-
+
–
– +
68.56
14.28 5.9 kNm
120 kN E
B C
RB = 17.14 RC = 17.14
4 m 3 m
75x10x3
2a 2
a2 = 500
mx 52
131
2B211A LMLLMA2LM
2
22
1
11 66
L
xa
L
xa
2MA x 10 + MB x 10 10
55006 xx
20MA + 10MB = - 1500
MA + 0.5MB = - 75 ------ (1)
Applying 3 – moment equation for A, B and C.
2211 2 LMLLMLM CBA 2
22
1
11
L
xa6
L
xa6
MA x 10 + 2MB (10 + 7) + MC x 7 10
5x500x6
7
)8.439(6
10MA + 34MB + 7MC = - 1123.02 ------ (2)
MA + 3.4MB + 0.7MC = - 112.303 ------ (2)
75x10x3
2a1
a1 = 500
mx 51
2xa 2
56.68x4x
2
1
4x
3
13
46.51x3x
2
1
3x
3
2
= - 439.80
132
Applying 3 – moment equation between B, C and C.
2C21C1B LMLLM2LM 2
22
1
11
L
xa6
L
xa6
7MB x 14MC 7
29.206 x
7MB x 14MC = 17.39 ------ (3)
Using (1), (2) and (3)
MA +0.5MB = -75
(1) – (2)
2.9MB + 0.7MC = –37.303 x 20 ------ (4)
58MB + 14MC = –746.06
7MA + 14MC = –17.39
51MB = –728.67
kNmM B 287.14
kNm85.67MA
kNm9.5MC
11 xa
= –
8.68x4x
2
1
4x
3
2
46.51x3x
2
1
3x
3
14
= 20.29