STATICALLY INDETERMINATE STRUCTURES

INTRODUCTION

Generally the trusses are supported on (i) a hinged support and (ii) a roller

support. The reaction components of a hinged support are two (in horizontal

and vertical directions), and for a roller support reaction component is only

one (in vertical direction). For any truss, if the number of members forming

the truss is m and the number of joints is j, then for the frame to be rigid

m = 2 j – 3.

The equations of static equilibrium are sufficient to determine the internal

forces in members of a rigid truss. If members are less than 2j – 3, the

truss becomes a collapsible truss and if m > (2j – 3), the truss becomes a

redundant truss.

Equations from statics,

i.e. summation of forces in x- and y-directions, respectively, is zero and

summation of moments of forces about any point is zero, are not sufficient

to determine the internal forces in members of a redundant truss. A unit

load method is used to analyze forces in members of a redundant truss.

Structures used in bridges are generally redundant with multiple degree of

redundancy. The study of such structures can be done by numerical

methods; we will analyze the trusses with single degree of redundancy.

There are many redundant structures in which strain compatibility condition

can be used for analyzing the forces in members of trusses or frames.

0,0,0 Oyx MFF

ANALYSIS OF REDUNDANT FRAMES WITH STRAIN COMPATIBILITY CONDITION Let us consider a system of three wires of steel, brass, and aluminium supporting a rigid bar OABCD, which is hinged at end O. A load P is applied at end D of the bar as shown in Figure 1. Let’s say the internal forces in bars of steel, brass, and aluminium are FA, FB, and FC, respectively. Then from the conditions of static equilibrium:

P = FA + FB + FC - RO

Figure 1 (a) A rigid bar supported by three wires (b) Deformations in wires

Moreover, taking moments about O, where RO is reaction at hinged end O, 4aP = a × FA + 2a × FB + 3a × FC

or

4P = FA + 2FB + 3FC

There are four unknowns, i.e. RO, FA, FB, and FC, but only two equations, so

conditions of static equilibrium cannot analyze the forces in this

indeterminate structure. We have to consider the deformation produced in

each bar after the application of load, or in other words strain compatibility

condition has to be used. The rigid bar is going to take the new position

OA’ B’ C’ D’ . From this diagram,

AA’ = δA deformation in bar A

BB’ = δB deformation in bar B

CC’ = δC deformation in bar C

but from the strain compatibility,

as is obvious from Figure 1 (b).

δB = 2 δA δC = 3 δA

where

From strain compatibility condition

Let us take Esteel = 2Ebrass = 3Eal

So, FB = FA

and

FC = FA

FB = FA and FC = FA

Putting these values in 4P = FA + 2FB + 3FC

From P = FA + FB + FC - RO , reaction

R0 reaction is downward.

PPPPFR AO 3

233

Example 7.1 A frame ABCD is shown in the figure. There are three bars AD, BD, and CD hinged at D as shown. Bars AD and CD are of copper while bar BD is of steel. Length of middle bar BD is L. Area of cross-section of each bar is the same. A load P is applied at end D. Determine the forces in the three bars. Given Esteel = 2Ecopper.

From the condition of static equilibrium:

P = FBD + FAD cos45 + FCD cos45 (i)

Due to symmetry FAD = FCD, because both

bars are identical and made of copper.

So,

P = FBD + 2FAD cos45

= FBD + 1.414FAD (ii) FAD

FBD

FCD

We have to use conditions of strain compatibility. After the application of load

P, point D moves down to D’ and all the bars are extended.

Extension in AD is D”D’ and extension in BD is DD’.

But DD′ sin45 = D” D′ (iii)

and

where A is the area of the cross-section.

Using Eq. (iii),

From Eqs. (iv) and (v),

Force

Now from Eq. (ii)

FAD

FBD

FCD

P = FBD + 1.414 FAD

= (0.7388 + 1.414 × 0.1847) P

= (0.7388 + 0.2612) P

= P

Let us verify the equilibrium of forces at joint D

FAD

FBD

FCD

Exercise 7.1

A rigid bar is suspended by three bars of steel and aluminium as

shown in the figure. The bars are of equal length and equal area of

cross-section. A load W is suspended on rigid bar so that the rigid bar

remains horizontal after the application of load. Determine forces in

aluminium and steel bars.

Esteel = 3Ealuminium

DEGREE OF REDUNDANCY

The total degree of redundancy or degree of indeterminacy of a frame is

equal to the number by which the unknown reaction components exceed the

condition of equation of equilibrium. The excess members are called

redundants.

Total degree of redundancy, T T = m – (2 j – R)

where m = total number of members in a frame,

j = total number of joints in frame, and

R = total number of reaction components.

Following criteria are applied for reaction components at supports of the frame: 1. For a roller support—reaction component is 1. 2. For a hinged support—reaction components are 2. 3. For a fixed support as in cantilever—reaction components are 3.

Let us consider the following example:

A frame ABCDEF shown in Figure 2 is

hinged at end A and roller supported at

end E. There are 11 members in frame.

Figure 2 Redundant frame

Number of members, m = 11

Number of joints, j = 6

Reaction components,

R = 2 (for hinged support at A) + 1 (for roller support at E) = 3

Degree of redundancy,

T = m – (2j – R)

= 11 – (2 × 6 – 3) = 11 – 9

= 2

Similarly, let us consider another frame ABCD, as

shown in Figure 3, having six members and four

joints. Frame is roller supported at A and hinged at

end C. Let us determine its degree of redundancy.

Figure 3 Triangular frame

m = 6, number of members

j = 4, number of joints

R = number of reaction components

= 2 (for hinged end C) + 1 (for roller support A)

= 3

Degree of redundancy

T = m − (2j − R) = 6 −(2×4−3)

= 6 − 5 = 1

Exercise 7.2

Determine the degree of redundancy for the frames shown in the figure.

[For all the cases, degree of redundancy is 1].

T = m – (2j – R)

ANALYSIS OF STATICALLY INDETERMINATE TRUSSES Method of virtual work or unit load method is one of the several methods available for solution of forces in members of an indeterminate truss. The following relationship is used to calculate the displacement in a member subjected to tensile or compressive load.

where S = internal force in truss members due to applied external loads

(external loads on truss plus a unit load)

U = axial load in truss members due to unit load applied externally at

a point,

E = Young’s modulus of elasticity of member,

L = length of member, and

A = area of cross-section of member.

Displacement

AE

SUL

We will consider trusses with only single degree of redundancy. The following steps are taken in order to determine the forces in members of an indeterminate truss. From the indeterminate truss, remove the redundant member so as to obtain a statically determinate structure as shown in Figures 4 (a) and (b). Obtain the forces in members of the statically determinate truss as shown in Figure 4(b) by equilibrium equations of statics.

Figure 4 (a) Statically indeterminate (b) Member BD removed statically determinate

Say these forces are S0 in each member.

Consider the same truss with one redundant member BD cut. Let the force Xa = 1 is applied on this member (from both ends, no support). Find the forces in members without the external force P (Figure 5).

Figure 5 Unit load along BD

Net force in each member S = S0 + Xaua Now let Xa be different from unity.

Let the relative displacement in redundant member be δa. Say the area of cross-section of each member is A and Young’s modulus is E.

or

Put (rigid structure) and get the value of Xa , or

or

Member S Magnitude of S

AB −0.707Xa + 0.5P

BC −P − 0.707Xa −0.5P

CD −P − 0.707Xa −0.5P

DA −0.707Xa + 0.5P

BD Xa −0.707P

AC + 0.707P

Forces in members of truss (putting the value of Xa)

Figure 6 shows the forces in the members of the truss.

Figure 6 Forces in members of truss

Equilibrium of Truss at joints: Let us check the equilibrium of forces at joints of the truss. For that, we have to first calculate the reactions. Taking moments about A, P× a = RD×a

Reaction, RD = P ↑

Reaction at A, RAH =

RAV = P ↓ (vertical component)

(horizontal component) P

Exercise 7.2 A redundant frame ABCD is shown in the figure. Material of all the bars is the same and the area of cross-section is also the same. Determine forces in all the members of the frame.

Example 7.3 For the truss shown in the figure, determine the support reactions and forces in all members. Area of cross-section of members AB, BC, and BD is 25 cm2 and for members AD and DC it is 22.5 cm2. Material is the same for all the members.

Number of members, m = 5

Number of joints, j = 4

Both the supports are hinged. So, the number of reaction components, R = 2 × 2 = 4

Degree of redundancy

T = m–(2j–R)

= 5–(2 × 4 – 4)

= 1

Vertical reactions at A and C RAV = RCV = 100 kN (due to symmetry). Let us say horizontal reaction at A and C is Xa. Remove Xa from both ends and calculate forces in members. Due to symmetry, we will solve for half the truss.

s = 45°

b = tan−1 0.5 = 26.56°

sin s = coss = 0.707

sin b = 0.447

cos b = 0.894

Angles,

FAB cos45 = FAD cos26.56 (x direction)

FAB × 0.707 = FAD × 0.894

FAB = 1.265 FAD

Forces RCV

Now putting the value of

FAD sin 26.56 + 100 = FAB sin 45

FAD × 0.447 + 100 = 0.707 FAB

FAB = 0.707 × (1.265 FAD)

= 0.894 FAD

Force,

(To have equilibrium of forces at joint A, FAD is tensile and FAB will be

compressive as shown).

FAB = FBC = –283 kN

Joint B

FAB sin 45 = FBC sin 45

But 283 × 0.707 × 2 = 200 + FBD

400 − 200 = FBD Force, FBD = +200 kN (tensile) Now applying the unit load Xa = 1 at ends A and C as shown in the figure.

Horizontal reactions

F′AD sin 26.56 = F′AB sin 45

F′AD × 0.447 = F′AB × 0.707

F′AD = 1.5816 F′AB (y direction)

F′AB × 0.707 + 1 = F′AD cos 26.56

= 0.894F′AD (x direction)

= 0.894 × 1.5816F′AB (putting the value of

F′AD)

= 1.414FAD

F′AB = + 1.414 (tension)

F′AD = −1.414 × 1.5816 = −2.237 (comp.)

A

F′BD = −2 × F′AB × 0.707 = −2 × 1.414 × 0.707

= −2 (compressive)

Joint B

S = S0 + Xaua

= Force in a member due to external load after removing

redundant reaction + force due to load Xa

Total force in a member,

Also,

If we tabulate the values of S, ua, L, A, etc., E is the same for all members:

Now

or

Reactions and forces in members

The forces are balanced at all the joints. All members are in compression.

Example 7.4

A triangular truss ABCD is shown in the figure. It is

subjected to a horizontal load P at joint B.

Determine the support reactions and forces in the

members of truss. The cross-sectional area of outer

members AB, BC, and CA is twice the cross-sectional

area of inner members AD, BD, and CD. The

material of all members is the same.

Taking moments of forces about point A

P × 0.866L = RCV × L

= 0.866P↑

RAV = 0.866P ↓

RAH = P as shown.

RCV, reaction

For degree of redundancy:

Number of members, m = 6

Number of joints, j = 4

Reactions components,

R = 2 (for hinged end) + 1 (roller supported end)

= 3

T = m − (2j − R)

= 6 − (2 ×4 − 3) = 1

Degree of redundancy,

Let us remove one member AC from the frame, and find forces S0 in members.

Member AC is removed (figure).

FBC cos 60° = FDC cos 30°

FBC × 0.5 = 0.866FDC

FBC = 1.732 FDC

Joint C

FDC sin 30° + 0.866P = FBC sin 60 °

0.5 FDC + 0.866P = 0.866 FBC

Moreover,

Putting the value of FBC

0.5 FDC + 0.866P = 0.866 × 1.732 FDC = 1.5 FDC Force,

FDC = 0.866P

FBC = 1.299 ÷ 0.866 = 1.5P

120°

Joint D ∠ ADC = ∠ CDB = ∠ BDA = 120° so FDC = FDB = FDA = 0.866P (tensile)

FAB cos60° + FAD cos30° = P

0.5FAB + 0.866 × 0.866P = P

0.5FAB = (P − 0.75P)

FAB = 0.5P

Joint A

FAB sin60° + FAD sin30° = 0.5 × 0.866P + 0.866 × 0.5P

= 0.866P (verified)

Checking for vertical loads.

120°

If we tabulate these values:

Now consider only Xa = 1 along member AC as shown in the figure.

F′AB sin60° = F′AD sin30°

F′AB × 0.866 = F′AD × 0.5

F′AD = 1.732F′AB

F′AD cos30° = F′AB cos60° + 1

F′AD × 0.866 = 0.5F′AB + 1

1.732 × 0.866F′AB = 0.5F′AB + 1

F′AB = 1 = F′BC

F′AD = −1.732 (comp)

= F′DC = F′DB

Putting the value of F′AD

If we tabulate these ua values also

Length AB = BC = CA = L, area of cross-section 2a

Material of all members is the same. Let us tabulate the different parameters.

Now

or

Resultant Forces in members

Example 7.5

Find the bending moment at any point of a semi-circular arch shown

in the figure. Both the supports are hinged.

Flexural rigidity EI is constant throughout.

For the arch, number of members, m = 1

number of joints, j = 2

number of reaction components,

R = 2 × 2= 4

(both ends are hinged).

Degree of redundancy,

T = m − (2j − R)

= 1 − (2 × 2 − 4) = 1

There will be reactions X and at each end as shown.

At a particular section a-a, bending moment

where k = –R sin θ.

Applying the principle of virtual work or unit load method,

Taking the advantage of symmetrical loading,

dsEI

M

dsX

M

EI

M

Xi

O

2

or

Therefore,

Bending moment at any section

Example 7.6 A rigid bar EF of negligible weight is suspended by four wires A, B, C, and D of the same length, same area of cross-section and same material. A load P is suspended at C as shown in the figure. Determine the forces in the four wires. From the conditions of static equilibrium,

P = PA + PB + PC + PD (i) Moments about point E PA × 0 + PB a + PC 2a + PD 3A = 2aP or 2P = PB + 2PC + 3PD (ii)

From the two equations, the values of four forces PA, PB, PC, and PD cannot

be determined. We have to take the help of a compatibility condition, i.e.

considering the deformation of the bar as shown in the figure.

Deformation in bar A =δA

Deformation in bar B = δA + δ

Deformation in bar C = δA + 2δ

Deformation in bar D = δA + 3δ

δA

Rigid bar EF will take the position A’B’C’D’

after the application of load.

Using Hooke’s law, in the elastic region,

deformation is directly proportional to load

applied.

Say δA ∝ PA (load on wire A)

δ ∝ P′ (load due to additional deformation δ)

δA A’

PB = PA + P′

PC = PA + 2P′

PD = PA + 3P′

So,

= PA + PB + PC + PD

= PA + PA + P′ + PA + 2P′ + PA + 3P′

= 4PA + 6P′

= P (iii)

Total reactions

2P = PA + P′ + 2 (PA + 2P′) + 3 (PA + 3P′)

2P = PA + 2PA + 3PA + P′ + 4P′ + 9P′

2P = 6PA + 14P′

Putting the values in (ii)

or P = 3PA + 7P′ (iv)

δA

PB = 2P′

PC = 3P′

PD = 4P′

From Eqs. (iii) and (iv),

4PA + 6P′ = 3PA + 7P′

PA = P′ (v)

So,

Total reaction

P′ + 2P′ + 3P′ + 4P′ = P

P′ = 0.10P (vi)

or forces in wire A, PA = P′ = 0.1P

in wire B, PB = 0.2P

in wire C, PC = 0.3P

in wire D, PD = 0.4P

δA

Exercise 7.3

A rigid bar EF of negligible weight is suspended by four wires A, B, C,

and D of the same material, same length and same area of cross-

section. A load P is suspended as shown in the figure. Determine the

forces in wires A, B, C, and D.