This session covers the basics that are required to analyse indeterminate trusses to maximum of two degree indeterminacy which includes,
strain energy stored due to axial loads and bending stresses,
maxwell's reciprocal deflection theorem,
Betti's law,
castigliano's theorems,
problems based on castigliano's theorems on beams and frames,
unit load method,
problems on trusses with unit load method,
lack of fit in trusses,
temperature effect on truss members.
Some basic defintions of the topics used in Strength of Materials subject. Pictorial presentation is more than details. Many examples are provided as well.
This document gives the class notes of Unit-8: Torsion of circular shafts and elastic stability of columns. Subject: Mechanics of materials.
Syllabus contest is as per VTU, Belagavi, India.
Notes Compiled By: Hareesha N Gowda, Assistant Professor, DSCE, Bengaluru-78.
This document gives the class notes of Unit 2 stresses in composite sections. Subject: Mechanics of materials.
Syllabus contest is as per VTU, Belagavi, India.
Notes Compiled By: Hareesha N Gowda, Assistant Professor, DSCE, Bengaluru-78.
Some basic defintions of the topics used in Strength of Materials subject. Pictorial presentation is more than details. Many examples are provided as well.
This document gives the class notes of Unit-8: Torsion of circular shafts and elastic stability of columns. Subject: Mechanics of materials.
Syllabus contest is as per VTU, Belagavi, India.
Notes Compiled By: Hareesha N Gowda, Assistant Professor, DSCE, Bengaluru-78.
This document gives the class notes of Unit 2 stresses in composite sections. Subject: Mechanics of materials.
Syllabus contest is as per VTU, Belagavi, India.
Notes Compiled By: Hareesha N Gowda, Assistant Professor, DSCE, Bengaluru-78.
Like Comment and Download if u like this presentation
Motion and deformation of material under action of
Force
Temperature change
Phase change
Other external or internal agents
These changes lead us to some properties that are called Mechanical properties
Some of the Mechanical Properties
Ductility
Hardness
Impact resistance
Fracture toughness
Elasticity
Fatigue strength
Endurance limit
Creep resistance
Strength of material
Ductility: ductility is a solid material's ability to deform under tensile stress
Hardness of a material may refer to resistance to bending, scratching, abrasion or cutting.
Impact resistance is the ability of a material to withstand a high force or shock applied to it over a short period of time
Plasticity: ability of a material to deform permanently by the
CONTENT:
1. Elastic strain energy
2. Strain energy due to gradual loading
3. Strain energy due to sudden loading
4. Strain energy due to impact loading
5. Strain energy due to shock loading
6. Strain energy due to shear loading
7. Strain energy due to bending (flexure)
8. Strain energy due to torsion
9. Examples
When a body is subjected to gradual, sudden or impact load, the body deforms and work is done upon it. If the elastic limit is not exceed, this work is stored in the body. This work done or energy stored in the body is called strain energy.
When a body is subjected to gradual, sudden or impact load, the body deforms and work is done upon it. If the elastic limit is not exceed, this work is stored in the body. This work done or energy stored in the body is called strain energy.
Lecture slides on the calculation of the bending stress in case of unsymmetrical bending. The Mohr's circle is used to determine the principal second moments of area.
This document gives the class notes of Unit 5 shear force and bending moment in beams. Subject: Mechanics of materials.
Syllabus contest is as per VTU, Belagavi, India.
Notes Compiled By: Hareesha N Gowda, Assistant Professor, DSCE, Bengaluru-78.
Like Comment and Download if u like this presentation
Motion and deformation of material under action of
Force
Temperature change
Phase change
Other external or internal agents
These changes lead us to some properties that are called Mechanical properties
Some of the Mechanical Properties
Ductility
Hardness
Impact resistance
Fracture toughness
Elasticity
Fatigue strength
Endurance limit
Creep resistance
Strength of material
Ductility: ductility is a solid material's ability to deform under tensile stress
Hardness of a material may refer to resistance to bending, scratching, abrasion or cutting.
Impact resistance is the ability of a material to withstand a high force or shock applied to it over a short period of time
Plasticity: ability of a material to deform permanently by the
CONTENT:
1. Elastic strain energy
2. Strain energy due to gradual loading
3. Strain energy due to sudden loading
4. Strain energy due to impact loading
5. Strain energy due to shock loading
6. Strain energy due to shear loading
7. Strain energy due to bending (flexure)
8. Strain energy due to torsion
9. Examples
When a body is subjected to gradual, sudden or impact load, the body deforms and work is done upon it. If the elastic limit is not exceed, this work is stored in the body. This work done or energy stored in the body is called strain energy.
When a body is subjected to gradual, sudden or impact load, the body deforms and work is done upon it. If the elastic limit is not exceed, this work is stored in the body. This work done or energy stored in the body is called strain energy.
Lecture slides on the calculation of the bending stress in case of unsymmetrical bending. The Mohr's circle is used to determine the principal second moments of area.
This document gives the class notes of Unit 5 shear force and bending moment in beams. Subject: Mechanics of materials.
Syllabus contest is as per VTU, Belagavi, India.
Notes Compiled By: Hareesha N Gowda, Assistant Professor, DSCE, Bengaluru-78.
This Slide explains basic theories in electrostatics, i.e. Coulomb's law, Electric field, electric potential, electric dipole, electric field due to electric dipole, etc.
Visit: https://phystudypoint.blogspot.com
Operation “Blue Star” is the only event in the history of Independent India where the state went into war with its own people. Even after about 40 years it is not clear if it was culmination of states anger over people of the region, a political game of power or start of dictatorial chapter in the democratic setup.
The people of Punjab felt alienated from main stream due to denial of their just demands during a long democratic struggle since independence. As it happen all over the word, it led to militant struggle with great loss of lives of military, police and civilian personnel. Killing of Indira Gandhi and massacre of innocent Sikhs in Delhi and other India cities was also associated with this movement.
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Safalta Digital marketing institute in Noida, provide complete applications that encompass a huge range of virtual advertising and marketing additives, which includes search engine optimization, virtual communication advertising, pay-per-click on marketing, content material advertising, internet analytics, and greater. These university courses are designed for students who possess a comprehensive understanding of virtual marketing strategies and attributes.Safalta Digital Marketing Institute in Noida is a first choice for young individuals or students who are looking to start their careers in the field of digital advertising. The institute gives specialized courses designed and certification.
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2. • Concept of strain energy
• Strain energy stored due to axial forces
• Strain energy stored due to bending stresses.
• Maxwell’s reciprocal deflection theorem or law of reciprocal
deflections
• Betti’s law
• Castigliano’s theorem-I
• Castigliano’s theorem-II
• Unit load method.
• Analysis of trusses having redundancy 1 and more than 1
• Lack of fit in indeterminate frames
• Temperature stresses in redundant frames
2v shiva
3. STRAIN ENERGY
• When an elastic member is deformed under the action
of an external loading the member is said to have
possessed or stored energy which is called the strain
energy of the member or the resilience of the member.
• Strain energy stored by a member = work done by the
external forces to produce deformation.
• If the elastic limit is not exceeded the external work
is transformed into elastic strain energy, which is
recoverable as member returns to its original
condition as the load is removed.
3v shiva
4. STRAIN ENERGY STORED DUE TO AXIAL forces
An elastic member of
Length = l;
C/S area = A;
Subjected to external axial load W.
Extension due to load = δ;
External work done We = ½.W.δ
strain energy stored = Wi;
External work done = strain energy stored. (We = Wi)
Let tension in member = S
For equilibrium,W = S;
Tensile stress, σ = S/A;
ε = σ/E;
δ = S.l/A.E
We = 1/2 W.δ = 1/2(S.S.l/AE)= S2l/2A.E
We = S2l/2A.E. (strain energy stored due to axial
loading on the member).
Strain energy stored per unit volume of the member
= (S2l/2A.E)/A.l
= S2/2A2.E = σ2/2E
4v shiva
5. • Strain energy stored due to bending:
• Consider a beam subjected to uniform moment M.
• Consider elemental length ds of the beam between two
sections 1-1 and 2-2.
• In this elemental length consider infinite number of
elemental cylinders of length ds and C/S area da.
• Intensity of stress in elemental cylinder, σ = M.y/I
• Energy stored by the elemental cylinder = energy stored
per unit volume X volume of the cylinder.
• = (σ2/2E) X ds.da
• = (M2 y2/2 I2E) X ds.da
• = Σ(M2 y2/2 I2E) X ds.da
• = (M2 ds / 2 I2E) X Σda .y2
• energy stored by ds length of beam = (M2 ds / 2 IE)
• Total energy stored by whole beam = ∫(M2 / 2 IE)ds
5v shiva
6. Find the deflection of the free end of a cantilever carrying a
concentrated load P at the free end
EI
pl
EI
lp
p
p
EI
lp
dx
EI
xp
l
3
6
.
2
1
.
2
1
loadexternalby thedonework
62
3
32
32
0
22
=
=
=
=
=
EI
dxM
2
Wi,cantileverbystoredenergystrain
2
6v shiva
7. Find the deflection of a simply supported beam carrying a
concentrated load P at mid-span
EI
pl
EI
lp
p
p
EI
lp
dx
EI
xp
l
48
96
.
2
1
.
2
1
loadexternalby thedonework
962
2
.
2
3
32
322/
0
2
=
=
=
=
=
=
EI
dxM
2
Wibeam,by wholestoredenergystrain
2
7v shiva
8. A portal frame ABCD has its end A hinged while the end D is placed on
rollers. A horizontal force P is applied on the end D as shown fig. determine
movement of D. Assume all members have the same flexural rigidity.
)32(
3
)32(
6
.
2
1
.
2
1
loadexternalby thedonework
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2
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bh
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ph
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hp
p
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ph
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bh
+=
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=
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+=
+=
=
8v shiva
9. Work done by a force on a member
• If a load acts on a member and
produces a deflection δ in its
line of action by virtue of its
own direct action, the work
done by the load W = ½.W.δ.
• If a member subjected to a
load K is given a deformation y
in the line of action of K by
virtue of some other external
agency, the work done by the
load K due to the above
deformation = Ky.
9v shiva
10. MAXWELL’S RECIPROCAL DEFLECTION THEOREM OR LAW
OF RECIPROCAL DEFLECTIONS
• James Clerk Maxwell, a Cavendish professor at Cambridge
University offered this law in 1864.
• In any beam or truss the deflection at any point D due to a load W at any
other point C is the same as the deflection at C due to the same load W
applied at D.
• Fig(i) shows a structure AB carrying load W at C. Due to this load,
deflection at point C = ∆c and
deflection at any other point D = ∆d
• Work done on the structure = 1/2.W. ∆c 10v shiva
11. • Now fig(ii) shows the same structure AB loaded with same
load(W) at points C and D.
• Now there will be further deflection of δc and δd at C and
D respectively.
• Total work done = 1/2.W.∆c + 1/2.W.δd + W.δc
11v shiva
12. • Fig(iii) shows a structure AB carrying load W at D.
Due to this load, deflection at point D = δd and
deflection at any other point C = δc
• Work done on the structure = 1/2.W. δd
12v shiva
13. • Now fig(iv) shows the same structure AB loaded
with same load(W) at points C and D.
• Now there will be further deflection of ∆c and ∆d at
C and D respectively.
• Total work done = 1/2.W.δd + 1/2.W.∆c + W.∆d
13v shiva
14. • Equating the two expressions obtained for the total work done when
both the loads are present on the structure, we have
= 1/2.W.∆c + 1/2.W.δd + W.δc = 1/2.W.δd + 1/2.W.∆c + W.∆d
W.δc = W.∆d
∴ δc = ∆d
i.e., The deflection at C due to the load W at D = The deflection at D
due to the same load W at C.
14v shiva
15. Maxwell’s law in the most general form is stated as follows;
δc = ∆d
In any structure whose material is elastic and obeys Hooke’s law and in
which the supports remain unyielding and the temperature remains
unchanged, the deflection of any point 1 in a direction ab due to a load
W acting at a point 2 acting in a direction cd is numerically equal to the
deflection of the point 2 in the direction cd due to load W acting at the
point 1 in the direction ab.
15v shiva
16. BETTI’S LAW
This is generalized form of Maxwell’s law offered by
Italian professor, E.Betti in 1872. The law is stated as
follows:
In any structure whose material is elastic and obeys Hooke’s
law and in which the supports remain unyielding and the
temperature remains constant, the virtual work done by a
system of forces P1, P2, P3 …. during the distortion caused
by a system of forces Q1, Q2, Q3 … is equal to the virtual
work done by the system of forces Q1, Q2, Q3 …during the
distortion caused by the system of forces, P1, P2, P3 …
16v shiva
17. In any structure whose materialis elastic
and obeys Hooke’s law and in which the
supports remain unyielding and the
temperatureremains constant, the virtual
work done by a system of forces P1, P2, P3
…. during the distortion caused by a system
of forces Q1, Q2, Q3 … is equal to the
virtual work done by the system of forces
Q1, Q2, Q3 …during the distortion caused
by the system of forces, P1, P2, P3 …
P1y1 + P2y2 + P3y3 = Q1Y1 + Q2Y2
+ Q3Y3
17v shiva
18. THE FIRST THEOREM OF CASTIGLIANO
• The first theorem of Castigliano states that the
partial derivative of total strain energy in any
structure with respect to the applied force or
moment gives the displacement or rotation
respectively at the point of application of force or
moment in the direction of applied force or
moment.
18v shiva
19. • Let the deflection at r be yr.
• We = external work done by the given load system.
• Wi = correspondingstrain energy stored.
∴We = Wi
Let ∆Pr aloneis applied getting
deformation of ∆yr.
19v shiva
21. • In case the deflection is to be determined at a point
where no load acts introduce a fictitious force Q at
the point in the direction of the desired deflection.
The partial derivative of strain energy stored
with respect to Q is determined, and Q is put equal
to zero.
21v shiva
22. Example: A simply supported beam carries a point load P eccentrically on
the span. Find the deflection under the load. Assume uniform flexural rigidity
• Reaction at A = Va = Pb/l
• Reaction at B = Vb = Pa/l
• The strain energy stored by the beam,
•
=
=
=
=
+
=
+
=
.
3
6
2bygivenisPloadunder thedeflection
6
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6
22
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2
22
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222
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EI
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xPb
EI
dxM
ba
22v shiva
23. The bend ABC shown in fig. carries a concentrated vertical load P at A. find
the vertical and horizontal deflections of A. Assume uniform flexural rigidity
( )
)3(
3
P,respect towithstoredenergystraintotaltheatedifferenti
A,ofdeflectionverticalthefindTo
)3(
6
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ha
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Pa
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Wi
ha
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dxM
v
ha
+
==
=
+
=
+=
23v shiva
24. ( ) ( )
( )
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QhPah
EI
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yQyPa
Q
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h
h
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=
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+=
++=
=
++==
24v shiva
25. THE SECOND THEOREM OF CASTIGLIANO
The second theorem of castigliano states that the work done by
external forces in a structure will be minimum. The theorem is
very useful in analysis of indeterminate structures.
Let W be the work done by the external forces on the structure, U be
the strain energy stored in the structure and W1 be the work done
by the reactive forces.
Strain energy U = W + W1
W = U – W1
By castigliano’s second theorem W should be minimum. Thus the
partial derivative of the work done with respect to external
forces will be zero.
In case the supports are unyielding the work done by reactive forces
will be zero. Strain energy stored will be equal to the work done by
external forces and will be the minimum. Thus the partial
derivative of strain energy with respect to redundant reaction will
be zero.
25v shiva
32. steps to analyze the redundant frame
Step-I: consider a member(say BC) as redundant and remove BC and find
member forces P1,P2,P3,.. For the given loading system.
Step-II: Remove the given loading systemand apply a pair of unit forces at B
and C. find the member forces K1,K2,K3,..
Step-III: Tabulate the above results as follows:
Step-IV: find a correcting factor X given by
member P K L A PKL/A K2 L/A
TOTAL Σ(PKL/A) Σ(K2 L/A)
=
A
LK
A
LKP
X
.
..
2
32v shiva
33. Step-V: Now find the force in any member by the relation
S = P + X.K
33v shiva
34. Example: Determine the forces in the members of the frame shown in
fig. sectional areas in millimeters
• Solution: Let BD is redundant, so remove
BD and find forces.
If we resolve and solve we get
force in members as;
Pca = 15√41KN(compressive)
Pcd = 60KN(Tensile)
Pda = 75KN(Tensile)
Pcb = 0
Pba = 0
34v shiva
35. • Now remove the given load and apply a pair of unit loads at B
and D in place of member BD.
Kdc = 4/√41(compressive)
Kda = 5/√41(compressive)
Kcb = 5/√41(compressive)
Kba = 4/√41(compressive)
Kca = 1(Tensile)
The actual force in the
member is given by
S = P + X.K,
=
A
LK
A
LKP
X
.
..
2
35v shiva
36. member P K L A PKL/A K2 L/A
AB 0 4/√41 4000 900 0 1.7344
BC 0 5/√41 5000 1200 0 2.5407
CD -60 4/√41 4000 900 -166.585 1.7344
DA -75 5/√41 5000 1200 -244.022 2.5407
AC 15√41 -1 1000√41 1500 -410 4.2687
BD 0 -1 1000√41 1500 0 4.2687
TOTAL -820.607 17.0876
36v shiva
37. • X = 820.607/17.0876
∴The actual force in the members are calculated as below;
• Sab = Pab + X.Kab = 0 + 48.0235X4/√41 = 30KN(compressive)
• Sbc = Pbc + X.Kbc = 0 + 48.0235X5/√41 = 37.5KN(compressive)
• Scd = Pcd + X.Kcd = -60 + 48.0235X4/√41 = -30KN (tensile)
• Sda = Pda + X.Kda = -75 + 48.0235X5/√41 = -37.5KN(tensile)
• Sac = Pac + X.Kac = 48.0235(-1)+15√41 =
48.023KN(compressive)
• Sbd = Pbd + X.Kbd = 0 + 48.0235(-1)= -48.023KN(tensile)
37v shiva
38. Frame with two redundant members
• Fig.1 shows a truss with two redundant members.
• Let S1,S2,S3,… be the actual forces.
• Let BF and DF be the redundant members.
• Let P1,P2,P3,..be the member forces after removing
BF and DF shown in fig.2.
• Let the member forces be K1,K2,K3,..due to unit load
at B and F as shown in fig.3
• Let the member forces be K’1,K’2,K’3,..due to unit load
at D and F as shown in fig.4
• The actual forces in the members of the truss are
given by,
• S1=P1+XK1+YK’1;
• S2=P2+XK2+YK’3; and so on…
38v shiva
40. Find the forces in the members FD and DH of the frame shown in
figure. The ratio of length to the cross-sectional area for all the
members is the same
• Given l/A is constant for all members.
• Consider FD and DH to be redundant members and assume them
to be removed.
• Let,
• X = Force in member FD
• Y = force in member DH
40v shiva
50. LACK OF FIT:
• Lack of fit is an occurance in trussesin which there is a differencebetween the
lengthof a member and the distance between the nodes it is supposedto fit.
• This happens because the connecting member is either too long or too short as
compared to the distance between corresponding nodes.
• It must also be noted here that lack of fit occurs only in indeterminate trusses because in
determinate trusses the variation in member length automatically adjusts the spacing
between the nodes, hence, lack of fit cannot occur.
• A lack of fit may be unintentionally introduced in a truss due to fabrication errors or it
may be deliberately introduced as a part of the design by the engineer.
• In cases of lack of fit, the member is either forcibly lengthened or forcibly shortened to
fit the truss depending on whether it is too short or too long respectively. This causes
the various members meeting at these joints to develop stresses of opposite nature. This
pre-stress occurring due to lack of fit may be used advantageously in design of truss.
• The members that develop tension due to loading may be introduced with a
compressive pre-stress and the members that develop compression may be introduced
with a tensile pre-stress. This effectively increases the stress range over which a
member may be loaded and the maximum load carrying capacity of the truss.
50v shiva
51. Consider a statically indeterminateframe in which
one of the members BC is shorterthan its length
by amount δ.
To fit this member in position, the joints will have to
be brought closer and member itself will get
elongated and thus will be in tension.
Let ‘x’ be the force in the member having lack of fit
To analyze the frame, it is made determinateby
removing the member having lack of fit
Unit forces are applied at the joints of the members
having lack of fit as shown in fig
• Let ‘k’ be the force in the member due to
unit forces.
• The force in the member due to force ‘x’will be
‘kx’.
• The strain energy in all the members of
determinateframe will be
51v shiva
54. • If any member in a redundant
frame is subjected to change in
temperature, there will be
change in the length of the
member and this will produce
stress in all the other members
of the frame.
• Consider figure(a),
• Let member BC be subjected to
temperature rise of T°C.
• Change in length of member
due to temperature change will
be
δl = loᾱT.
54v shiva
55. • The member subjected to
temperature change is removed and
the frame is made determinate.
• Unit loads are applied at the joints of
the members as shown in figure(b)
and forces in various members
calculated.
• Let the force in a member due to unit
loads be k.
• The force in the member due to force
‘X’ in redundant member BC will be
kX.
• Total strain energy in all the
members of determinate frame will
be
55v shiva
57. A triangularframeshown in figurehas sectionalarea equal to 35X10-4 m2 for all the
members. The member AB was found to be 1X10-3 m shorterthan its correctlengthat
the timeof assembling.Find the forces in the members, if the member AB is forced in
position.Take Young’smodulusas 2.1X108 KN/m2
Given:
Indeterminatestructureby degree of static
indeterminacy = 1
area of members = 35X10-4 m2
Error in length(δ)= 1X10-3 m
Young’s modulus, E = 2.1X108 KN/m2
Let T = force in member AB due to lack of fit.
Remove member AB and apply unit tensile force
at jointsA and B as shown in figure(b)
57v shiva
59. member Length in
m
AREA E(KN/m2) k K
2
l/AE Actual force
kT in KN
AB 10 0.0035 210000000 1 1.36054E-05 -8.97
BC 10 0.0035 210000000 1 1.36054E-05 -8.97
AC 10 0.0035 210000000 1 1.36054E-05 -8.97
CO 5.77 0.0035 210000000 -1.732 2.35496E-05 15.53604
BO 5.77 0.0035 210000000 -1.732 2.35496E-05 15.53604
AO 5.77 0.0035 210000000 -1.732 2.35496E-05 15.53604
TOTAL 0.000111465
59v shiva
61. Problem:In the planebraced frameshown in figure(a)all the membershave same
cross sectionalarea of 8 cm2 and are made of same material. The member AC in the
framewas initiallyshort by 0.25 cm. determinetheforce in each member.
In the aboveframedeterminetheforce in each member when member AC is subjected
to a risein temperatureof 20°C.
ᾱ = 1.1 x 10-5 /°C and E = 2 x 105 N/mm2
61v shiva
62. • Given:
• Area, A = 8 cm2 = 0.0008m2
• E = constant= 2 x 105 N/mm2 = 2 x 1011 N/m2
• ᾱ = 1.1 x 10-5 /°C
• Error in length of member AC = 0.25 cm = 0.0025m
• Member AC is removed and unit loadsare applied as shown in figure(b).
• Forces in the members are calculatedas,
62v shiva
64. memb
er L A E K K2L/AE
Force in
member due
to lack of
fit(KX)
Force in
member
due to
temperat
ure(KX1)
Final
force
AB 4 0.0008 200000000 -0.707
1.24962E-
05 14.64197 -7.2821 7.35987
BC 4 0.0008 200000000 -0.707
1.24962E-
05 14.64197 -7.2821 7.35987
CD 4 0.0008 200000000 -0.707
1.24962E-
05 14.64197 -7.2821 7.35987
DA 4 0.0008 200000000 -0.707
1.24962E-
05 14.64197 -7.2821 7.35987
BD 5.656 0.0008 200000000 1 0.00003535 -20.71 10.3 -10.41
FB 2.828 0.0008 200000000 0 0 0 0 0
FC 2.828 0.0008 200000000 0 0 0 0 0
EA 2.828 0.0008 200000000 0 0 0 0 0
ED 2.828 0.0008 200000000 0 0 0 0 0
AC 5.6535 0.0008 200000000 1
3.53344E-
05 -20.71 10.3 -10.41
TOTAL
0.00012066
9
64v shiva
66. The diagonal AC of the frame work, shown in figure was found
to be 0.625x10-3 m longer than its correct length, at the time of
assembling. The member AC was forced in position. Find the
forces induced in the members of the frame.
Take E = 2X108 KN/m2 and diameter of each member =
2X10-2 m
66v shiva