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UNIT-I

INDETERMINATE FRAMES

Introduction:

Plane frames are also made up of beams and columns, the only difference being they

are rigidly connected at the joints as shown in the Fig.

Major portion of this course is devoted to evaluation of forces in frames for variety of

loading conditions.

Internal forces at any cross section of the plane frame member are: bending moment,

shear force and axial force.

As against plane frame, space frames members may be oriented in any direction. In

this case, there is no restriction of how loads are applied on the space frame.

Equations of Static Equilibrium

Static Indeterminacy

To evaluate the external reactions, the deformed shape and internal stresses in the

structure.

If this can be accomplished by equations of equilibrium, then such structures are

known as determinate structures.

However, in many structures it is not possible to determine either reactions or internal

stresses or both using equilibrium equations alone.

The indeterminacy in a structure may be external, internal or both.

A structure is said to be externally indeterminate if the number of reactions exceeds

the number of equilibrium equations.

Beams shown in Fig.(a) and (b) have four reaction components, whereas we have

only 3 equations of equilibrium. Hence the beams in Figs. (a) and (b) are externally

indeterminate to the first degree.

Similarly, the beam and frame shown in Figs. (c) and (d) are externally indeterminate

to the 3rd

degree.

Truss:

In these structures, reactions could be evaluated based on the equations of

equilibrium.

However, member forces cannot be determined based on statics alone.

In Fig. (a), if one of the diagonal members is removed (cut) from the structure then

the forces in the members can be calculated based on equations of equilibrium.

Thus, structures shown in Figs. (a) and (b) are internally indeterminate to first degree.

The truss and frame shown in Figure (a) and (b) are both externally and internally

indeterminate.

Consider a planar stable truss structure having m members and j joints.

Let the number of unknown reaction components in the structure be r .

Now, the total number of unknowns in the structure is m + r .

At each joint we could write two equilibrium equations for planar truss structure, viz.,

Σ Fx = 0 and Σ Fy = 0.

Hence total number of equations that could be written is 2j.

If 2 j = m + r, then the structure is statically determinate as the number of unknowns

are equal to the number of equations available to calculate them.

The degree of indeterminacy may be calculated as

i = (m + r) − 2 j

Example:

Plane Frame

For example, the plane frame has 15 members, 12 joints and 9 reaction components.

Hence, the degree of indeterminacy of the structure is

i = (15× 3 + 9) −12 × 3 = 18

Energy Methods

1. Castigliano’s theorem

Castigliano’s first theorem is being used in structural analysis for finding deflection of

an elastic structure based on strain energy of the structure.

The Castigliano’s theorem can be applied when the supports of the structure are

unyielding and the temperature of the structure is constant.

Castigliano’s First Theorem

For linearly elastic structure, where external forces only cause deformations, the

complementary energy is equal to the strain energy.

For such structures, the Castigliano’s first theorem may be stated as the first partial

derivative of the strain energy of the structure with respect to any particular force

gives the displacement of the point of application of that force in the direction of its

line of action.

Let P1 , P2,...., Pn be the forces acting at from the left end on a simply supported

beam of span L.

u1 ,u2 ... un be the displacements at the loading points P1 , P2,...., Pn respectively as

shown in Figure.

Now, assume that the material obeys Hooke’s law and invoking the principle of

superposition, the work done by the external forces is given by

Work done by the external forces is stored in the structure as strain energy in a

conservative system.

Hence, the strain energy of the structure is,

Displacement u1 below point P1 is due to the action of P1 , P2,...., Pn acting at

distances x1, x2, ….. xn respectively from left support. Hence, u1 may be expressed as,

u1 = a11 P1 + a12 P2 + ………….+ a1n Pn

In general,

Problem:

Find the vertical deflection at A of the structure shown Figure. Assume the flexural

rigidity EI and torsional rigidity GJ to be constant for the structure.

Solution:

The beam segment BC is subjected to bending moment Px (0 < x < a ; x is

measured from C) and the beam element AB is subjected to torsional moment of magnitude

Pa and a bending moment of Px ( 0 ≤ x ≤ b ; x is measured from B) . The strain energy stored

in the beam ABC is,

After simplifications

Vertical deflection uA at A is,

Castigliano’s Second Theorem

In any elastic structure having ‘n’ independent displacements u1, u2, …. un

corresponding to external forces P1, P2, …..Pn along their lines of action, if strain energy is

expressed in terms of displacements then equilibrium equations may be written as follows.

2. Theorem of Least Work

The partial derivative of strain energy of a statically indeterminate structure with

respect to statically indeterminate action should vanish as it is the function of such

redundant forces to prevent any displacement at its point of application.

The forces developed in a redundant framework are such that the total internal strain

energy is a minimum.

Consider a beam that is fixed at left end and roller supported at right end as shown in

Figure.

Let P1, P2, …..Pn be the forces acting at distances x1, x2, …. xn from the left end of the

beam of span L .

Let u1, u2, …. un be the displacements at the loading points P1, P2, …..Pn respectively

as shown in Fig. (a).

This is a statically indeterminate structure and choosing Ra as the redundant reaction,

we obtain a simple cantilever beam as shown in Fig. (b).

Invoking the principle of superposition, this may be treated as the superposition of

two cases,

viz, a cantilever beam with loads P1, P2, …..Pn and a cantilever beam with

redundant force Ra

In the first case (a), obtain deflection below A due to applied loads P1, P2, …..Pn.

This can be easily accomplished through Castigliano’s first theorem.

Since there is no load applied at A , apply a fictitious load Q at A as in Fig. 2.

Let ua be the deflection below A.

Now the strain energy Us stored in the determinate structure (i.e. the support A

removed) is given by,

1. Find the horizontal deflection catch joint C of the Rigid frame shown in fig.

Take E=2×105

N/mm2, I=4×10

9mm

4

Step 1: Find the HA, RA and RC

∑HA = 0

-HA+10 = 0

HA = 10 KN

Take M@ A=0

(-RC×4) + (5×42/2) + (10×2) = 0

RC = 15 KN

Total upward force = Total downward force

RA+RC = (5×3)

RA + 15 = 20

RA = 5 KN

Step:2

Deflection ∆= 0∫2 (M1m1/EI)dx+0∫

2 (M2m2/EI)dx+0∫

4 (M3m3/EI)dx ------------- 1

Find M1, M2 & M3

M1 = HA×X1 + RA×0

= 10 X 1

M2 = HA× (2+X2) + (RA×0)-(10×X2)

= 20

M3=-RC×X3+ (5X32/2)

=-15X3+2.5X32

Step: 3

All given loads are removed and 1KN Horizontal force act @ point C.

Find the HA, RA and RC

∑HA = 0

-HA+1=0

HA =1 KN

Take M@ A=0

(-RC×4)+ (1×4) =0

RC=1 KN

Total upward force = Total downward force

RA+RC=0

RA+1=0

RA = -1 KN

Find m1, m2 & m3

m1 = HA×X1+RA×0

= X1

M2 = HA× (2+X2) + (RA×0)

= 2+X2

M3 = -RC×X3

= -X3

All values sub in eqn 1 , and we get

∆= 0∫2 ((10X1×X1)/EI)dx+0∫

2 (20×(2+X2)/EI)dx+0∫

4 ((-15x3+2.5x3

2)(-X3)/EI)dx

=10/EI 0∫2 X1

2 dx+1/EI0∫

2 (40+20X2)dx+1/EI0∫

4 ((15x3

2-2.5x3

3)/EI)dx

=10/EI ( 23/3) +1/EI(40×2+(20×2

2)/2)+1/EI ((15×4

3-2.5×4

4/4)/EI)

=306.67/EI

=306/(2×108×4×10

-3)

∆=3.83 mm

2. Find the horizontal deflection and slope at point D. Teke EI is a constant.

Step 1: Find the HA, RA and RC

∑HA=0

-HA+0=0

HA=0

Take M@ D=0

(RA×3)-(10×32/2)=0

RA=15 KN

Total upward force = Total downward force

RA+RD=(10×3)

RD+15=30

RD=15 KN

Deflection ∆= 0∫4 (M1m1/EI)dx+0∫

3 (M2m2/2EI)dx+0∫

4 (M3m3/EI)dx −−−1

Find M1, M2 & M3

M1=RA×0

=0

M2=(RA×X2)-(10×X22/2)

=15X2-5X22

M3=-RD×0

=0

Step:3

All given loads are removed and 1KN Horizontal force act @ point D.

Find the HA, RA and RC

∑HA=0

-HA+1=0

HA=1 KN

Take M@ D=0

(RA×3)+(HA×0)=0

RA=0 KN

Total upward force = Total downward force

RA+RD=0

RD+0=0

RD=0 KN

Find m1, m2 & m3

m1=RA×0+HA×X1

=X1

m2=RA×X2+(HA×4)

=4

M3=-RD×0-HA×X3

= -X3

All values sub in eqn 1 , and we get

∆= 0∫4 ((X1×0)/EI)dx+0∫

3 (15X2-5X2

2)(4)/2EI)dx+0∫

4 (0×X3)/EI)dx

=2/EI 0∫3 (15×X2-5X2

2 )dx

=2/EI ( (15×32/2)-(5×3

2/3

))dx

∆=45/EI

Step : 4

All given loads are removed and 1KNM Moment act @ point D

Find the HA, RA and RC

∑HA=0

-HA+0=0

HA=0 KN

Take M@ D=0

(RA×3)+1=0

RA=-0.33KN

Total upward force = Total downward force

-0.33+RD=0

RD=0.33 KN

Find m1, m2 & m3

m1=RA×0+HA×X1

=-0.33×0+0×X1

=0

m2=RA×X2+(HA×4)

=-0.33X2+0×4

=-0.33X2

M3=-RD×0+1

= 1

All values sub in eqn 1 , and we get

ⱷ = 0∫4 0 dx+0∫

3 (15X2-5X2

2)( -0.33X2)/2EI)dx+0∫

4 0dx

=1/2EI 0∫3 (1.65×X2

3-4.95X2

2 )dx

1/2EI (1.65×34/4-4.95×3

3 /3)

ⱷ =-5.57/EI

3. Determine the Deflection at joint B ss beam shown in fig by the principle of virtual

work. Take E=2×105

N/mm2, I=4×10

9mm

4

Step 1: Find RA and RD

Take M@ A=0

(RD×9)-(8×6)-(5×3)=0

RD=7 KN

Total upward force = Total downward force

RA+RD=13

RA=6 KN

Deflection ∆= 0∫3 (M1m1/EI)dx+0∫

3 (M2m2/2EI)dx+0∫

3 (M3m3/EI)dx −−−1

Find M1, M2 & M3

M1=RA×X1

=6X1

M2=RA(3+X2)-5X2

= 6(3+X2)-5X2

=X2+18

M3=-RD×X3

=-7X3

All given loads are removed and 1KN Vertical Force act @ point B

Take M@ A=0

(RD×9)-(1×3)=0

RD=0.33KN

Total upward force = Total downward force

-0.33+RA=1

RA=0.67 KN

Find m1, m2 & m3

m1=RA×X1

=-0.67×X1

m2=RA(3+X2)-(1×X2)

=-0.33X2+2.01

M3=-RD×X3

= -0.33X3

∆=1/EI 0∫3 (6X1×0.67X1)dx+1//2EI 0∫

3 (X2+18)(-0.33X2 +2.01)dx+

1//EI 0∫3 (-7X3)(-0.33X3 )dx

∆=101/EI

=101/(2×108×3×10

-3)

∆= 0.126 mm

4. Using the principle of virtual work. Determine the vertical deflection components of

joint C.Take E, A as a constant.

Step 1:

Joint D:

∑V=0

-10-FCD sin 45=0

-FCD sin 45=10

FCD=-14.14 KN ↓

∑H=0

-FED-FCD cos45=0

-FED=FCD cos 45

FED=10 KN ↑

Joint E:

∑V=0

-FCE=0

FCE = 0 KN

∑H=0

FDE-FAE=0

FAE = 10 KN ↑

Joint C:

∑V=0

FCE+FCD sin 45-FCBsin 45=0

0-14.14sin45-FBCsin45=0

-FBCsin45=10

FBC =-14.14 KN ↓

∑H=0

-FCF+FCDcos45-FBCcos45=0

-FCF-14.14cos45-(-14.14cos45)=0

FCE=0 KN

Step 2:

All loads are removed and 1 KN point load act @

point D

Joint D:

∑V=0

-1-KCD sin 45=0

-KCD sin 45=1

KCD=-1.14 KN ↓

∑H=0

-KED-KCD cos45=0

-KED=KCD cos 45

KED=1 KN ↑

Joint E:

∑V=0

-FCE=0

FCE = 0 KN

∑H=0

FDE-FAE=0

FAE = 10KN ↑

Joint C:

∑V=0

KCE+KCD sin 45-KCBsin 45=0

0-1.414sin45-KBCsin45=0

-KBCsin45=1↑

KBC =-1.414 KN ↓

∑H=0

-KCF+KCDcos45-KBCcos45=0

-KCF-1.414cos45-(-1.414cos45)=0

FCE=0 KN

Member F

KN

K

KN

A

m2

E

KN/m2

L

m

FKL/AE

m

AE 10 1 A E 3 30/AE

ED 10 1 A E 3 30/AE

EC 0 0 A E 3 0

FC 0 0 A E 3 0

BC -14.14 -1.414 A E 4.24 84.77/AE

CD -14.14 -1.414 A E 4.24 84.77/AE

Deflection =∑FKL/AE

=230/AE ‘m’

5. Using the principle of virtual work.Determine the vertical and Horizontal deflection

components of joint C.Take E=200×106

KN/m2,.

And sectional area of each bar

A=100×10-6

m2.

Solution:

Sin ⱷ =OPP/HYP=CB/2

CB=2×sin30

CB=1 m

Sin 60 =AC/2

AC=1.73 m

M@ B=0

(RA×2)-(100×0.5)=0

RA=25 KN

RA+RB=100 KN

RB=75 KN

Joint A:

∑V=0

RA+FAC sin30=0

25+FAC sin30=0

FAC=-50KN ↓

∑H=0

FAB+FAC cos30=0

FAB-50cos30=0

FAB=43.3 KN ↑

Joint B:

∑V=0

75+FBC sin60=0

75+FBC sin60=0

FBC=-86.6KN ↓

(i) Vertical deflection

All given loads are removed and 1 KN point load act

@ point C.

RA×2-1×0.5=0

RA=0.25 KN

RB=0.75 KN

Joint A:

∑V=0

RA+KAC sin30=0

0.25+KAC sin30=0

KAC=-0.5KN ↓

∑H=0

KAB+KAC cos30=0

KAB-0.5cos30=0

KAB=0.433 KN ↑

Joint B:

∑V=0

0.75+KBC sin60=0

0.75+KBC sin60=0

KBC=-0.866KN ↓

(ii)Horizontal deflection

All given loads are removed and 1 KN

Horizontal load act @ point C

M@B=0

RA×2+1×0.866+HA×0=0

RA=-0.433 KN

M@A=0

-RB×2+1×0.866=0

RB=0.433 KN

∑HA=0

-HA+1=0

HA=1 KN

Joint A:

∑V=0

RA+KAC sin30=0

0.433+KAC sin30=0

KAC=0.866 KN ↑

∑H=0

KAB+KAC cos30=0

KAB+0.866cos30=0

KAB=0.25KN ↑

Joint B:

∑V=0

0.433+KBC sin60=0

0.433+KBC sin60=0

KBC=-0.5 KN ↓

(i) VERTICAL DEFLECTION

Member F KN K KN L m

E kn/m2

A m2 FKL/AE m

AB 43.3 0.433 2 100×106 100×10

-6 1.87

AC -50 -0.5 1.73 100×106 100×10

-6 2.18

BC -86.60 -0.866 1 100×106 100×10

-6 3.74

∆=∑FKL/AE=7.79 mm

(ii) HORIZONTAL DEFLECTION

Member F KN K KN L m

E kn/m2

A m2 FKL/AE m

AB 43.3 0.25 2 100×106 100×10

-6 1.08

AC -50 0.866 1.73 100×106 100×10

-6 -3.74

BC -86.60 -0.5 1 100×106 100×10

-6 2.16

∆=∑FKL/AE=-0.505 mm