STATICALLY INDETERMINATE STRUCTURES - …kisi.deu.edu.tr/emine.cinar/ASM16-Statically...ANALYSIS OF...
Transcript of STATICALLY INDETERMINATE STRUCTURES - …kisi.deu.edu.tr/emine.cinar/ASM16-Statically...ANALYSIS OF...
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.
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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.