elements of mechnaical engineering full syllabus
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About This Presentation
civil engineering bec 1st year
Slide Content
BEC-I : Elements of Mechanical
Engineering (MECI-101)
(Tutorial)
Dr B R AMBEDKAR NATIONAL INSTITUTE OF TECHNOLOGY
JALANDHAR
DEPARTMENT OF MECHANICAL ENGINEERING
2/2/2023 1
Engineering (MECI-101)
(Tutorial)
Dr B R AMBEDKAR NATIONAL INSTITUTE OF TECHNOLOGY
JALANDHAR
DEPARTMENT OF MECHANICAL ENGINEERING
2/2/2023 1
SolidMechanicsIntroduction:Systemofforces,coplanarconcurrentforce
system,compositionandresolutionofforce,equilibriumofrigidbodies,
freebodydiagram,Lami’stheorem.
Analysisofframedstructure:Reactioninbeamwithdifferentend
conditions,determinationofreactionsinmembersoftrusses:a)Analytical
methodsb)Graphicalmethod
Centreofgravityandmomentofinertia:ConceptofC.Gandcentroid,
positionofcentroid,theoremofparallelandperpendicularaxes,moment
ofinertiaofsimplegeometricalfigures.
Stressandstrain:Conceptofstressandstrain,simplestresses,tensile,
compressive,shear,bendingandtorsion,stress-straincurves,elongation
ofbars,compositebars,thermalstresses,elasticconstants.
Elements of Mechanical Engineering (MECI-101)
Syllabus (Part-I)
2/2/2023 2
system,compositionandresolutionofforce,equilibriumofrigidbodies,
freebodydiagram,Lami’stheorem.
Analysisofframedstructure:Reactioninbeamwithdifferentend
conditions,determinationofreactionsinmembersoftrusses:a)Analytical
methodsb)Graphicalmethod
Centreofgravityandmomentofinertia:ConceptofC.Gandcentroid,
positionofcentroid,theoremofparallelandperpendicularaxes,moment
ofinertiaofsimplegeometricalfigures.
Stressandstrain:Conceptofstressandstrain,simplestresses,tensile,
compressive,shear,bendingandtorsion,stress-straincurves,elongation
ofbars,compositebars,thermalstresses,elasticconstants.
Elements of Mechanical Engineering (MECI-101)
Syllabus (Part-I)
2/2/2023 2
2/2/2023
3
Solid Mechanics Introduction
Force –
Forceisanexternalagencywhichchangesortendsto
changethestateofrestoruniformlinearmotionofthe
body.
Characteristics of a Force -
—Magnitude
—Point of application
—Line of action
—Direction
COMPOSITION AND RESOLUTION OF VECTORS
1. Parallelogram Law of Vectors
2. Triangle Law of Vectors
3. Polygon Law of Forces (Vectors)
Resultant forces
R =
Resolution of Vectors
F
x
θ
F
x
= F cos θ
F
y
= F sinθ
Component of Force F
3
Solid Mechanics Introduction
Force –
Forceisanexternalagencywhichchangesortendsto
changethestateofrestoruniformlinearmotionofthe
body.
Characteristics of a Force -
—Magnitude
—Point of application
—Line of action
—Direction
COMPOSITION AND RESOLUTION OF VECTORS
1. Parallelogram Law of Vectors
2. Triangle Law of Vectors
3. Polygon Law of Forces (Vectors)
Resultant forces
R =
Resolution of Vectors
F
x
θ
F
x
= F cos θ
F
y
= F sinθ
Component of Force F
2/2/2023 4
Solid Mechanics Introduction
Question 1. A black weighing W = 10 kN is resting on an inclined plane as shown in Fig. (a). Determine its components
normal to and parallel to the inclined plane
Solution:Theplanemakesanangleof20°tothehorizontal.Hencethenormaltotheplanemakesananglesof70°to
thehorizontali.e.,20°tothevertical[Ref.Fig].IfABrepresentsthegivenforceWtosomescale,ACrepresentsits
componentnormaltotheplaneandCBrepresentsitscomponentparalleltotheplane.
Thus from ∆ ABC,
Component normal to the plane = AC
= W cos 20°
= 10 cos 20°= 9.4 kN
(b) Component parallel to the plane = W sin 20°
= 10 sin 20°
= 3.42 kN, down the plane
Free body diagram
Solid Mechanics Introduction
Question 1. A black weighing W = 10 kN is resting on an inclined plane as shown in Fig. (a). Determine its components
normal to and parallel to the inclined plane
Solution:Theplanemakesanangleof20°tothehorizontal.Hencethenormaltotheplanemakesananglesof70°to
thehorizontali.e.,20°tothevertical[Ref.Fig].IfABrepresentsthegivenforceWtosomescale,ACrepresentsits
componentnormaltotheplaneandCBrepresentsitscomponentparalleltotheplane.
Thus from ∆ ABC,
Component normal to the plane = AC
= W cos 20°
= 10 cos 20°= 9.4 kN
(b) Component parallel to the plane = W sin 20°
= 10 sin 20°
= 3.42 kN, down the plane
Free body diagram
2/2/2023 5
Solid Mechanics Introduction
Question2.Theresultantoftwoforces,oneofwhichisdoubletheotheris260N.Ifthedirectionofthelargerforceis
reversedandtheotherremainunaltered,themagnitudeoftheresultantreducesto180N.Determinethemagnitudeof
theforcesandtheanglebetweentheforces.
Given that -
Solution :
F
1
= F,F
2
= 2F and
θ be the angle between the two forces
Case 1 :
(1)
Case 2 :
5F
2
–4F
2
cos θ= 32400(2)
Adding equation (i) and (ii), we get
F = 100 N
F
1
= F = 100 N,
F
2
= 2F = 200 N
Substituting the values of F
1
and F
2
in eqn. (1), we get,
Solid Mechanics Introduction
Question2.Theresultantoftwoforces,oneofwhichisdoubletheotheris260N.Ifthedirectionofthelargerforceis
reversedandtheotherremainunaltered,themagnitudeoftheresultantreducesto180N.Determinethemagnitudeof
theforcesandtheanglebetweentheforces.
Given that -
Solution :
F
1
= F,F
2
= 2F and
θ be the angle between the two forces
Case 1 :
(1)
Case 2 :
5F
2
–4F
2
cos θ= 32400(2)
Adding equation (i) and (ii), we get
F = 100 N
F
1
= F = 100 N,
F
2
= 2F = 200 N
Substituting the values of F
1
and F
2
in eqn. (1), we get,
2/2/2023 6
Question 1: The resultant of two forces one of which is 3 times the other is 300 N. When the direction of smaller force is
reversed, the resultant is 200 N. Determine the two forces and the angle between them.
[Ans. F1 = 80.6 N, F2 = 241.8 N, θ = 50.13°]
Question 2: A rocket is released from a fighter plane at an angle upward 20°to the vertical with an acceleration of 8
m/sec
2
. The gravitational acceleration is 9.1 m/sec
2
downward. Determine the instantaneous acceleration of the rocket when
it was fired.
[Ans. 9.849 m/sec
2
, θ= 49.75°to vertical]
Question 1: The resultant of two forces one of which is 3 times the other is 300 N. When the direction of smaller force is
reversed, the resultant is 200 N. Determine the two forces and the angle between them.
[Ans. F1 = 80.6 N, F2 = 241.8 N, θ = 50.13°]
Question 2: A rocket is released from a fighter plane at an angle upward 20°to the vertical with an acceleration of 8
m/sec
2
. The gravitational acceleration is 9.1 m/sec
2
downward. Determine the instantaneous acceleration of the rocket when
it was fired.
[Ans. 9.849 m/sec
2
, θ= 49.75°to vertical]
2/2/2023 7
Solid Mechanics Introduction
VARIGNON’S THEOREM
Thealgebraicsumofthemomentsofasystemofcoplanarforcesaboutamomentcentreintheirplaneisequaltothe
momentoftheirresultantforceaboutthesamemomentcentre.
Question 3: Find the moment of 100 N force acting at B about point A as shown in Fig.
Solution:
Momentof100NforceaboutthepointAisequaltosumofthemomentsofits
componentsaboutA.
100Nforcemayberesolvedintoitshorizontalcomponentsas100cos60°and
verticalcomponent100sin60°.
From Varignon’s theorem,
Taking clockwise moment as positive,
M
A
= 100 cos 60°×500 –100 sin 60°×400
= 25, 000 –34, 641.02
= –9641.02 N-mm
M
A
= 9641.016 N-mm Anticlockwise
Solid Mechanics Introduction
VARIGNON’S THEOREM
Thealgebraicsumofthemomentsofasystemofcoplanarforcesaboutamomentcentreintheirplaneisequaltothe
momentoftheirresultantforceaboutthesamemomentcentre.
Question 3: Find the moment of 100 N force acting at B about point A as shown in Fig.
Solution:
Momentof100NforceaboutthepointAisequaltosumofthemomentsofits
componentsaboutA.
100Nforcemayberesolvedintoitshorizontalcomponentsas100cos60°and
verticalcomponent100sin60°.
From Varignon’s theorem,
Taking clockwise moment as positive,
M
A
= 100 cos 60°×500 –100 sin 60°×400
= 25, 000 –34, 641.02
= –9641.02 N-mm
M
A
= 9641.016 N-mm Anticlockwise
8
Solid Mechanics Introduction
COMPOSITION OF CONCURRENT COPLANAR FORCES
LetF
1
,F
2
,F
3
andF
4
showninFig.(a)bethesystemoffour
forcestheresultantofwhichisrequired.
The procedure to get the resultant -
Step 1:Find the components of all the forces in X and Y directions.
Thus, F
1x
, F
2x
, F
3x
, F
4x
, F
1y
, F
2y
, F
3y
, and F
4y
, are obtained.
Step 2:FindthealgebraicsumofthecomponentforcesinXand
Ydirections.
ΣF
x
= F
1x
+ F
2x
+ F
3x
+ F
4x
ΣF
y
= F
1y
+ F
2y
+ F
3y
+ F
4y
Step 3:Nowthesystemofforcesisequaltotwo
mutuallyperpendicularforces,namely,ΣF
x
and
ΣF
y
asshowninFig.(b)
Solid Mechanics Introduction
COMPOSITION OF CONCURRENT COPLANAR FORCES
LetF
1
,F
2
,F
3
andF
4
showninFig.(a)bethesystemoffour
forcestheresultantofwhichisrequired.
The procedure to get the resultant -
Step 1:Find the components of all the forces in X and Y directions.
Thus, F
1x
, F
2x
, F
3x
, F
4x
, F
1y
, F
2y
, F
3y
, and F
4y
, are obtained.
Step 2:FindthealgebraicsumofthecomponentforcesinXand
Ydirections.
ΣF
x
= F
1x
+ F
2x
+ F
3x
+ F
4x
ΣF
y
= F
1y
+ F
2y
+ F
3y
+ F
4y
Step 3:Nowthesystemofforcesisequaltotwo
mutuallyperpendicularforces,namely,ΣF
x
and
ΣF
y
asshowninFig.(b)
9
Solid Mechanics Introduction
Question 4: Determine the resultant of the three forces acting on a hook as shown in Fig.
Solution:
Force (N) x component (N)y component (N)
70 45 53.62
80 72.5 33.81
50 35.36 -35.36
ΣF
x
= 152.86Σ F
y
= 52.07
Given that-
R = 161.48 N
α = 18.81°
Solid Mechanics Introduction
Question 4: Determine the resultant of the three forces acting on a hook as shown in Fig.
Solution:
Force (N) x component (N)y component (N)
70 45 53.62
80 72.5 33.81
50 35.36 -35.36
ΣF
x
= 152.86Σ F
y
= 52.07
Given that-
R = 161.48 N
α = 18.81°
2/2/2023 10
Solid Mechanics Introduction
Question 5: A system of forces acting on a body resting on an inclined plane is as shown in Fig. Determine the resultant force
if θ = 60°and if W = 1000 N; N = 500 N; F = 100 N; and T = 1200 N.
Solution:
Inthisproblem,notethatselectingXandYaxesparalleltotheplaneand
perpendiculartotheplaneisconvenient.
R
x
= ΣF
x
= T –F –W sin θ
= 1200 –100 –1000 sin 60°= 233.97 N
R
y
= ΣF
y
= N –W cos 60°= 500 –1000 cos 60°
= 0
∴Resultant is force of 233.97 N directed up the plane
Solid Mechanics Introduction
Question 5: A system of forces acting on a body resting on an inclined plane is as shown in Fig. Determine the resultant force
if θ = 60°and if W = 1000 N; N = 500 N; F = 100 N; and T = 1200 N.
Solution:
Inthisproblem,notethatselectingXandYaxesparalleltotheplaneand
perpendiculartotheplaneisconvenient.
R
x
= ΣF
x
= T –F –W sin θ
= 1200 –100 –1000 sin 60°= 233.97 N
R
y
= ΣF
y
= N –W cos 60°= 500 –1000 cos 60°
= 0
∴Resultant is force of 233.97 N directed up the plane
2/2/2023 11
Question 1: Three forces acting at a point are shown in Fig. The direction of the 300 N forces may vary, but the angle
between them is always 40°. Determine the value of θ for which the resultant of the three forces is directed parallel to b-b.
[Ans. θ = 6.35°]
Question 1: Three forces acting at a point are shown in Fig. The direction of the 300 N forces may vary, but the angle
between them is always 40°. Determine the value of θ for which the resultant of the three forces is directed parallel to b-b.
[Ans. θ = 6.35°]
12
Solid Mechanics Introduction
EQUILIBRIANT OF A FORCE SYSTEM
The force which brings the body to the state of equilibrium and obviously, this forces is equal in magnitude, but opposite in
the direction to the resultant.
COMPOSITION OF COPLANAR NON-CONCURRENT FORCE SYSTEM
ΣF
x
–algebraic sum of the components of all forces in x direction
ΣF
y
–algebraic sum of the components of all forces in y direction
Α–inclination of the resultant R to x direction
ΣM
O
–algebraic sum of the moments of all the forces about point O
d –is distance of the resultant R from the point O.
Where,
x AND y INTERCEPTS OF RESULTANT
Solid Mechanics Introduction
EQUILIBRIANT OF A FORCE SYSTEM
The force which brings the body to the state of equilibrium and obviously, this forces is equal in magnitude, but opposite in
the direction to the resultant.
COMPOSITION OF COPLANAR NON-CONCURRENT FORCE SYSTEM
ΣF
x
–algebraic sum of the components of all forces in x direction
ΣF
y
–algebraic sum of the components of all forces in y direction
Α–inclination of the resultant R to x direction
ΣM
O
–algebraic sum of the moments of all the forces about point O
d –is distance of the resultant R from the point O.
Where,
x AND y INTERCEPTS OF RESULTANT
13
Solid Mechanics Introduction
Question 6: Find the resultant of the force system shown in Fig. acting on a lamina of equilateral triangular shape
Solution:
ΣF
x
= 80 –100 cos 60°–120 cos 30°= –73.92 N
R
x
= 73.92 N
ΣF
y
= 80 + 120 sin 30°–100 sin 60°
R
y
= 53.40 N
R = 91.19 N
α = 35.84°
Let x be the distance from A at which the resultant cuts AC.
Then taking A as moment centre
53.40x = 80 ×100 sin 60°+ 80 ×50 + 120 sin 30°×100
x = 317.008 mm to the right of A
Solid Mechanics Introduction
Question 6: Find the resultant of the force system shown in Fig. acting on a lamina of equilateral triangular shape
Solution:
ΣF
x
= 80 –100 cos 60°–120 cos 30°= –73.92 N
R
x
= 73.92 N
ΣF
y
= 80 + 120 sin 30°–100 sin 60°
R
y
= 53.40 N
R = 91.19 N
α = 35.84°
Let x be the distance from A at which the resultant cuts AC.
Then taking A as moment centre
53.40x = 80 ×100 sin 60°+ 80 ×50 + 120 sin 30°×100
x = 317.008 mm to the right of A
14
Question 6: Four forces having magnitudes of 200 N, 400 N, 600 N and 800 N respectively, are acting along the four sides (1
m each) of a square ABCD taken in order, as shown in Fig. Determine the magnitude and direction of the resultant force.
Solution:
ΣF
x
= 200 –600 = –400 N = 400 N
ΣF
y
= 400 –800 = –400 N = 400 N
= 565.68 N
Let x be the distance from A along x axis, where resultant cuts
AB. Then
ΣM
A
= 400 ×1 + 600 ×1 = 1000 N-m
Solid Mechanics Introduction
Question 6: Four forces having magnitudes of 200 N, 400 N, 600 N and 800 N respectively, are acting along the four sides (1
m each) of a square ABCD taken in order, as shown in Fig. Determine the magnitude and direction of the resultant force.
Solution:
ΣF
x
= 200 –600 = –400 N = 400 N
ΣF
y
= 400 –800 = –400 N = 400 N
= 565.68 N
Let x be the distance from A along x axis, where resultant cuts
AB. Then
ΣM
A
= 400 ×1 + 600 ×1 = 1000 N-m
Solid Mechanics Introduction
15
Question 6: Determine the resultant of system of parallel forces acting on a beam as shown in Fig.
Solution:
ΣF
x
= 0 N
ΣF
y
= 80 –30 + 40 –50 + 60 = 100 kN
= 100 kN
Let x be the distance from A along x axis, where resultant cuts x
axis. Then
ΣM
A
= 80 ×0 –30 ×2 + 40 ×4 –50 ×8 + 60 ×10
= 300 kN-m
Solid Mechanics Introduction
Question 6: Determine the resultant of system of parallel forces acting on a beam as shown in Fig.
Solution:
ΣF
x
= 0 N
ΣF
y
= 80 –30 + 40 –50 + 60 = 100 kN
= 100 kN
Let x be the distance from A along x axis, where resultant cuts x
axis. Then
ΣM
A
= 80 ×0 –30 ×2 + 40 ×4 –50 ×8 + 60 ×10
= 300 kN-m
Solid Mechanics Introduction
2/2/2023 16
FREE BODY DIAGRAM
Adiagramofthebodyinwhichthebodyunder
considerationisfreedfromallthecontactsurfaces
andshowsalltheforcesactingonit(including
reactionsatcontactsurfaces),iscalledaFreeBody
Diagram(FBD).
EQUILIBRIUM OF BODIES
•Thealgebraicsumofthecomponentofforces
alongeachofthetwomutuallyperpendicular
directionsiszero(translatorymotioniszero).
•Thealgebraicsumofmomentofalltheforces
aboutanypointintheplaneiszero(rotational
momentiszero)
Σ??????
?
= 0; (X direction)
Σ??????
?
= 0; (Y direction)
ΣM = 0
Mathematically -
FREE BODY DIAGRAM
Adiagramofthebodyinwhichthebodyunder
considerationisfreedfromallthecontactsurfaces
andshowsalltheforcesactingonit(including
reactionsatcontactsurfaces),iscalledaFreeBody
Diagram(FBD).
EQUILIBRIUM OF BODIES
•Thealgebraicsumofthecomponentofforces
alongeachofthetwomutuallyperpendicular
directionsiszero(translatorymotioniszero).
•Thealgebraicsumofmomentofalltheforces
aboutanypointintheplaneiszero(rotational
momentiszero)
Σ??????
?
= 0; (X direction)
Σ??????
?
= 0; (Y direction)
ΣM = 0
Mathematically -
2/2/2023 17
Solid Mechanics Introduction
EQUILIBRIUM OF CONCURRENT FORCE SYSTEMS
Lami’s theorem :
If a body is in equilibrium under the action of three forces, each force is proportional to the sine of the angle between the
other two forces.
Solid Mechanics Introduction
EQUILIBRIUM OF CONCURRENT FORCE SYSTEMS
Lami’s theorem :
If a body is in equilibrium under the action of three forces, each force is proportional to the sine of the angle between the
other two forces.
2/2/2023 18
Solid Mechanics Introduction
Question 7. Determine the horizontal force P to be applied to a block of weight 1500 N to hold it in position on a smooth
inclined plane AB which makes an angle of 30°with the horizontal.
Solution:
Method 1:
ΣF
y
= 0,
R cos 30°–1500 = 0
R = 1732.06 N
ΣF
x
= 0,
P –R sin 30°= 0
P = R sin 30°
P = 866.03 N
Method 2: Lami’s theorem
R = 1732.06 and P = 866.03
Solid Mechanics Introduction
Question 7. Determine the horizontal force P to be applied to a block of weight 1500 N to hold it in position on a smooth
inclined plane AB which makes an angle of 30°with the horizontal.
Solution:
Method 1:
ΣF
y
= 0,
R cos 30°–1500 = 0
R = 1732.06 N
ΣF
x
= 0,
P –R sin 30°= 0
P = R sin 30°
P = 866.03 N
Method 2: Lami’s theorem
R = 1732.06 and P = 866.03
2/2/2023 19
Solid Mechanics Introduction
Question8.AcordACB5mlongisattachedatpointsAandBtotwoverticalwalls3mapartasshowninFig.(a).A
pullyCofnegligibleradiuscarriesasuspendedloadof200Nandisfreetorollwithoutfrictionalongthecord.
Determinethepositionofequilibrium,asdefinedbythedistanceX,thatthepulleywillassumeandalsothetensile
forceinthecord.
Solution:
ΣH = 0
T cos θ
1
–T cos θ
2
= 0
∴θ
1
= θ
2
= θ
Now, let BC be extended to D
∆CFD = ∆CFA
∴CD = AC
BD = BC + CD = BC + AC = length of chord = 5 m DE = 3 m
∴BE = 4 m
As ∆BHI is similar to ∆BDE
AH = 3 –0.75 = 2.25
x = 1.125 m. Since AH = 2x
At C,
ΣV = 0
2 ×T sin θ = 200
2 ×T ×4 5 = 200
∴T = 125 N
Solid Mechanics Introduction
Question8.AcordACB5mlongisattachedatpointsAandBtotwoverticalwalls3mapartasshowninFig.(a).A
pullyCofnegligibleradiuscarriesasuspendedloadof200Nandisfreetorollwithoutfrictionalongthecord.
Determinethepositionofequilibrium,asdefinedbythedistanceX,thatthepulleywillassumeandalsothetensile
forceinthecord.
Solution:
ΣH = 0
T cos θ
1
–T cos θ
2
= 0
∴θ
1
= θ
2
= θ
Now, let BC be extended to D
∆CFD = ∆CFA
∴CD = AC
BD = BC + CD = BC + AC = length of chord = 5 m DE = 3 m
∴BE = 4 m
As ∆BHI is similar to ∆BDE
AH = 3 –0.75 = 2.25
x = 1.125 m. Since AH = 2x
At C,
ΣV = 0
2 ×T sin θ = 200
2 ×T ×4 5 = 200
∴T = 125 N
2/2/2023 20
Solid Mechanics Introduction
Question9.Arollerofradiusr=300mmandweight2000Nistobepulledoveracurbofheight150mm[Fig.]bya
horizontalforcePappliedtotheendofastringwoundtightlyaroundthecircumferenceoftheroller.Findthemagnitude
ofPrequiredtostarttherollermoveoverthecurb.WhatistheleastpullPthroughthecentreofthewheeltojustturn
therolleroverthecurb?
Solution:
Now in ∆ AOB,
∠OAB = ∠OBA
since OA = OB
but
∠OAB +∠OBA = α
2 ∠OBA= 60 °
∠OBA = 30°
ΣV = 0
R cos 30°–2000 = 0
R = 2309.40 N
ΣH = 0,
gives P –R sin 30°= 0
P = 2309.40 ×sin 30°
P = 1154.70 N.
Case 2:
Case 1:
From triangle of forces ABC,
we get P = AC = AB sin θ
= 2000 sin 60°
P = 1732.05 N
Solid Mechanics Introduction
Question9.Arollerofradiusr=300mmandweight2000Nistobepulledoveracurbofheight150mm[Fig.]bya
horizontalforcePappliedtotheendofastringwoundtightlyaroundthecircumferenceoftheroller.Findthemagnitude
ofPrequiredtostarttherollermoveoverthecurb.WhatistheleastpullPthroughthecentreofthewheeltojustturn
therolleroverthecurb?
Solution:
Now in ∆ AOB,
∠OAB = ∠OBA
since OA = OB
but
∠OAB +∠OBA = α
2 ∠OBA= 60 °
∠OBA = 30°
ΣV = 0
R cos 30°–2000 = 0
R = 2309.40 N
ΣH = 0,
gives P –R sin 30°= 0
P = 2309.40 ×sin 30°
P = 1154.70 N.
Case 2:
Case 1:
From triangle of forces ABC,
we get P = AC = AB sin θ
= 2000 sin 60°
P = 1732.05 N
2/2/2023 21
Solid Mechanics Introduction
Question 10. A system of connected flexible cables shown in Fig. (a) is supporting two vertical forces 200 N and 250 N at
points B and D. Determine the forces in various segments of the cable
Applying Lami’s theorem to the system of forces at point D
T
1
= 224.14 N
T
2
= 183.01 N
Consider the system of forces acting at B
ΣV = 0
T
3
cos 30°–200 –T
2
cos 60°= 0
T
3
= 336.60 N
ΣH = 0
T
4
–T
2
sin 60°–T
3
sin 30°= 0
T
4
= 183.01 ×sin 60°+ 336.60 sin 30°
T
4
= 326.79 N
Solution:
Solid Mechanics Introduction
Question 10. A system of connected flexible cables shown in Fig. (a) is supporting two vertical forces 200 N and 250 N at
points B and D. Determine the forces in various segments of the cable
Applying Lami’s theorem to the system of forces at point D
T
1
= 224.14 N
T
2
= 183.01 N
Consider the system of forces acting at B
ΣV = 0
T
3
cos 30°–200 –T
2
cos 60°= 0
T
3
= 336.60 N
ΣH = 0
T
4
–T
2
sin 60°–T
3
sin 30°= 0
T
4
= 183.01 ×sin 60°+ 336.60 sin 30°
T
4
= 326.79 N
Solution:
2/2/2023 22
Solid Mechanics Introduction
Question 10. Two smooth spheres each of radius 100 mm and weight 100 N, rest in a horizontal channel having vertical walls,
the distance between which is 360 mm. Find the reactions at the points of contacts A, B, C and D shown in Fig. (a)
Solution:
Free body diagram
Consider sphere No. 1
ΣV = 0
R
B
×0.6 = 100
R
B
= 166.67 N.
ΣH = 0,
R
A
= R
B
×0.8
∴R
A
= 133.33 N
Consider sphere No. 2
ΣV = 0,
R
c
= 100 + R
B
×0.6
∴R
c
= 200 N.
ΣH = 0,
R
D
= R
B
×0.8
∴R
D
= 133.33 N
Solid Mechanics Introduction
Question 10. Two smooth spheres each of radius 100 mm and weight 100 N, rest in a horizontal channel having vertical walls,
the distance between which is 360 mm. Find the reactions at the points of contacts A, B, C and D shown in Fig. (a)
Solution:
Free body diagram
Consider sphere No. 1
ΣV = 0
R
B
×0.6 = 100
R
B
= 166.67 N.
ΣH = 0,
R
A
= R
B
×0.8
∴R
A
= 133.33 N
Consider sphere No. 2
ΣV = 0,
R
c
= 100 + R
B
×0.6
∴R
c
= 200 N.
ΣH = 0,
R
D
= R
B
×0.8
∴R
D
= 133.33 N
Reaction in beam with different end conditions
TYPES OF BEAMS
BEAMS:Amemberwhichissubjectedtopredominantly
transverseloadsandsupportedinsuchawaythatrigidbody
motionispreventedisknownasbeam.Therearedifferent
typesofBEAMS.
a)
b) c)
a) Simply Supported Beam
b) Cantilever beam
c) Overhanging beam
2/2/2023 23
TYPES OF BEAMS
BEAMS:Amemberwhichissubjectedtopredominantly
transverseloadsandsupportedinsuchawaythatrigidbody
motionispreventedisknownasbeam.Therearedifferent
typesofBEAMS.
a)
b) c)
a) Simply Supported Beam
b) Cantilever beam
c) Overhanging beam
2/2/2023 23
Reaction in beam with different end conditions
1.ConcentratedLoads –Thisistheloadactingforverysmalllengthofthe
beam.(alsoknownaspointload,TotalloadWisactingatonepoint)
TYPES OF LOADS ON BEAMS
2.Uniformlydistributedload–Thisistheloadactingforaconsiderable
lengthofthebeamwithsameintensityofwkN/mthroughoutitsspread.Total
intensity,W=w×L(actsatL/2fromoneendofthespread)
Total intensity, W = w ×L (acts at L/2
from one end of the spread)
2/2/2023 24
1.ConcentratedLoads –Thisistheloadactingforverysmalllengthofthe
beam.(alsoknownaspointload,TotalloadWisactingatonepoint)
TYPES OF LOADS ON BEAMS
2.Uniformlydistributedload–Thisistheloadactingforaconsiderable
lengthofthebeamwithsameintensityofwkN/mthroughoutitsspread.Total
intensity,W=w×L(actsatL/2fromoneendofthespread)
Total intensity, W = w ×L (acts at L/2
from one end of the spread)
2/2/2023 24
Reaction in beam with different end conditions
TYPES OF LOADS ON BEAMS
3.Uniformlyvaryingload–Thisloadactsforaconsiderable
lengthofthebeamwithintensityvaryinglinearlyfrom‘0’at
oneendtowkN/mtotheotherrepresentingatriangular
distribution.
Total intensity of load = area of triangular
spread of the load W = 1/2×w ×L. (acts at
2×L/3 from ‘Zero’ load end)
2/2/2023 25
TYPES OF LOADS ON BEAMS
3.Uniformlyvaryingload–Thisloadactsforaconsiderable
lengthofthebeamwithintensityvaryinglinearlyfrom‘0’at
oneendtowkN/mtotheotherrepresentingatriangular
distribution.
Total intensity of load = area of triangular
spread of the load W = 1/2×w ×L. (acts at
2×L/3 from ‘Zero’ load end)
2/2/2023 25
Reaction in beam with different end conditions
TYPES OF SUPPORTS
Supports:Astructureissubjectedtoexternalforcesandtransfersthese
forcesthroughthesupportsontothefoundation.Thereforethesupport
reactionsandtheexternalforcestogetherkeepthestructurein
equilibrium.
There are different types of supports.
a)
b) c)
d) e)
a) Flexible cable, belt, Chain, rope
b) Smooth surfaces
c) Roller Support
d) Hinged or pinned support
e) Fixed or built in support
2/2/2023 26
TYPES OF SUPPORTS
Supports:Astructureissubjectedtoexternalforcesandtransfersthese
forcesthroughthesupportsontothefoundation.Thereforethesupport
reactionsandtheexternalforcestogetherkeepthestructurein
equilibrium.
There are different types of supports.
a)
b) c)
d) e)
a) Flexible cable, belt, Chain, rope
b) Smooth surfaces
c) Roller Support
d) Hinged or pinned support
e) Fixed or built in support
2/2/2023 26
Reaction in beam with different end conditions
Statically determinate beam
Usingtheequationsofequilibriumgiven
below,ifallthereactioncomponents
canbefoundout,thenthebeamisa
staticallydeterminatebeam
Σ??????
?
= 0; (X direction)
Σ??????
?
= 0; (Y direction)
ΣM = 0
Step3:Assumebeamisstaticallyinequilibrium
Σ??????
?
=0;(Xdirection)
Σ??????
?
=0;(Ydirection)
ΣM=0
Steps for solution:
Step 1: Name the reaction point (If not given in question)
Step 2: Convert all load into point load
2/2/2023 27
Statically determinate beam
Usingtheequationsofequilibriumgiven
below,ifallthereactioncomponents
canbefoundout,thenthebeamisa
staticallydeterminatebeam
Σ??????
?
= 0; (X direction)
Σ??????
?
= 0; (Y direction)
ΣM = 0
Step3:Assumebeamisstaticallyinequilibrium
Σ??????
?
=0;(Xdirection)
Σ??????
?
=0;(Ydirection)
ΣM=0
Steps for solution:
Step 1: Name the reaction point (If not given in question)
Step 2: Convert all load into point load
2/2/2023 27
Reaction in beam with different end conditions
Question1:ThebeamABofspan12mshowninFig(a)ishingedatAandisonrollersatB.Determinethereactionsat
AandBfortheloadingalsofindtheresultantforceatAshownintheFigure.
Solution:
Step 2:Step 1:
Step3: Σ??????
?
=0;(Xdirection)
??????
?
–15cos30°–20cos45°=0
??????
?
=27.1325kN
Σ??????
?
=0
??????
?
×12–10×4–15sin30°×6–20sin45°×10=0
??????
?
=18.8684kN
Σ??????
?
=0;(Ydirection)
??????
?
+18.8684–10–15sin30°–20sin45°=0
??????
?
=12.7737kN
Resultant force at A: ??????
?
= 29.989 kN. At α= 25.21°from x axis
2/2/2023 28
Question1:ThebeamABofspan12mshowninFig(a)ishingedatAandisonrollersatB.Determinethereactionsat
AandBfortheloadingalsofindtheresultantforceatAshownintheFigure.
Solution:
Step 2:Step 1:
Step3: Σ??????
?
=0;(Xdirection)
??????
?
–15cos30°–20cos45°=0
??????
?
=27.1325kN
Σ??????
?
=0
??????
?
×12–10×4–15sin30°×6–20sin45°×10=0
??????
?
=18.8684kN
Σ??????
?
=0;(Ydirection)
??????
?
+18.8684–10–15sin30°–20sin45°=0
??????
?
=12.7737kN
Resultant force at A: ??????
?
= 29.989 kN. At α= 25.21°from x axis
2/2/2023 28
Reaction in beam with different end conditions
Question2:FindthereactionsatsupportsAandBinthebeamABshowninFigure.
Solution:
Step 2:Step 1:
Step3: Σ??????
?
=0
??????
?
sin 60°×6 –60 sin 60°×1 –80 ×sin 75°×3 –50 ×sin 60°×5.5 = 0
∴ ??????
?
= 100.4475 kN
Σ??????
?
=0;(Xdirection)
??????
?
+ 60 cos 60°–80 cos 75°+ 50 cos 60°–RB cos 60°= 0
??????
?
= –60 cos 60°+ 80 cos 75°–50 cos 60°+ 100.4475 cos 60°
??????
?
= 15.9293 kN
Σ??????
?
=0;(Ydirection)
??????
?
+ ??????
?
sin 60°–60 sin 60°–80 sin 75°–50 sin 60°= 0
??????
?
= –100.4475 sin 60°+ 60 sin 60°+ 80 sin 75°+ 50 sin 60°
??????
?
= 85.5468 kN
Resultant force at A: ??????
?
= 87.0172 kN.At α= 79.45°from x axis
2/2/2023 29
Question2:FindthereactionsatsupportsAandBinthebeamABshowninFigure.
Solution:
Step 2:Step 1:
Step3: Σ??????
?
=0
??????
?
sin 60°×6 –60 sin 60°×1 –80 ×sin 75°×3 –50 ×sin 60°×5.5 = 0
∴ ??????
?
= 100.4475 kN
Σ??????
?
=0;(Xdirection)
??????
?
+ 60 cos 60°–80 cos 75°+ 50 cos 60°–RB cos 60°= 0
??????
?
= –60 cos 60°+ 80 cos 75°–50 cos 60°+ 100.4475 cos 60°
??????
?
= 15.9293 kN
Σ??????
?
=0;(Ydirection)
??????
?
+ ??????
?
sin 60°–60 sin 60°–80 sin 75°–50 sin 60°= 0
??????
?
= –100.4475 sin 60°+ 60 sin 60°+ 80 sin 75°+ 50 sin 60°
??????
?
= 85.5468 kN
Resultant force at A: ??????
?
= 87.0172 kN.At α= 79.45°from x axis
2/2/2023 29
Reaction in beam with different end conditions
Question3:ThecantilevershowninFigureisfixedatAandisfreeatB.Determinethereactionswhenitisloadedas
shownintheFigure
Solution:
Step 1
Step3: Σ??????
?
=0;(Xdirection)
??????
?
= 0 kN
Σ??????
?
=0;(Ydirection)
??????
?
–16 ×2 –20 –12 –10 = 0
∴??????
??????
= 74 kN.
Σ??????
?
=0
??????
?
–16 ×2 ×1 –20 ×2 –12 ×3 –10 ×4 = 0
∴ ??????
??????
= 148 kN-m
1m
1m
32kN
Step 2
2/2/2023 30
Question3:ThecantilevershowninFigureisfixedatAandisfreeatB.Determinethereactionswhenitisloadedas
shownintheFigure
Solution:
Step 1
Step3: Σ??????
?
=0;(Xdirection)
??????
?
= 0 kN
Σ??????
?
=0;(Ydirection)
??????
?
–16 ×2 –20 –12 –10 = 0
∴??????
??????
= 74 kN.
Σ??????
?
=0
??????
?
–16 ×2 ×1 –20 ×2 –12 ×3 –10 ×4 = 0
∴ ??????
??????
= 148 kN-m
1m
1m
32kN
Step 2
2/2/2023 30
Reaction in beam with different end conditions
Questionsfor practice
Question1:FindthereactionsatsupportsAandBoftheloadedbeamshowninFigure.
Step: 1
120 kN
2/2/2023 31
Σ M
B
= 0, gives
R
A
×9 –20 ×7 –30 ×4 ×5 –60 sin 45°×2 = 0
R
A
= 91.6503 kN
Σ H
A
= 0, gives
H
B
–60 cos 45°= 0
H
B
= 42.4264 kN
Σ V
A
= 0
91.6503 + V
B
–20 –30 4 –60 sin 45°= 0
V
B
= 90.7761 kN.
R
B
= (42.4264
2
+ 90.7761
2
)
1/2
R
B
= 100.2013 kN.
α = tan
-
(
=4.;::
86.86:8
)
α= 64.95°
Questionsfor practice
Question1:FindthereactionsatsupportsAandBoftheloadedbeamshowninFigure.
Step: 1
120 kN
2/2/2023 31
Σ M
B
= 0, gives
R
A
×9 –20 ×7 –30 ×4 ×5 –60 sin 45°×2 = 0
R
A
= 91.6503 kN
Σ H
A
= 0, gives
H
B
–60 cos 45°= 0
H
B
= 42.4264 kN
Σ V
A
= 0
91.6503 + V
B
–20 –30 4 –60 sin 45°= 0
V
B
= 90.7761 kN.
R
B
= (42.4264
2
+ 90.7761
2
)
1/2
R
B
= 100.2013 kN.
α = tan
-
(
=4.;::
86.86:8
)
α= 64.95°
2/2/2023 32
Reaction in beam with different end conditions
Questionsfor practice
Question2:FindthereactionsatsupportsAandBoftheloadedbeamshowninFigure.
40 kN
2/2/2023 33
ΣV = 0, gives
V
A
–20 ×2 –15 ×10 = 0
VA = 65 kN.
ΣM = 0, gives
MA –20 ×2 ×1 –15 ×3 –30 –10 ×5 = 0
MA = 165 kN-m
Questionsfor practice
Question2:FindthereactionsatsupportsAandBoftheloadedbeamshowninFigure.
40 kN
2/2/2023 33
ΣV = 0, gives
V
A
–20 ×2 –15 ×10 = 0
VA = 65 kN.
ΣM = 0, gives
MA –20 ×2 ×1 –15 ×3 –30 –10 ×5 = 0
MA = 165 kN-m
Reaction in beam with different end conditions
Questionsfor practice
Question 3: Determine the reactions at supports A and B of the overhanging beam shown in Figure.
60 kN
2/2/2023 34
ΣM
A
= 0, gives
R
B
×5 –30 ×1 –20 ×3 ×(2 + 1.5) –40 ×6.5 = 0
∴R
B
= 100 kN
Σ V = 0, gives
R
A
+ R
B
–30 –20 ×3 –40 = 0
Σ R
A
= 130 –R
B
= 130 –100
∴R
A
= 30 kN.
Questionsfor practice
Question 3: Determine the reactions at supports A and B of the overhanging beam shown in Figure.
60 kN
2/2/2023 34
ΣM
A
= 0, gives
R
B
×5 –30 ×1 –20 ×3 ×(2 + 1.5) –40 ×6.5 = 0
∴R
B
= 100 kN
Σ V = 0, gives
R
A
+ R
B
–30 –20 ×3 –40 = 0
Σ R
A
= 130 –R
B
= 130 –100
∴R
A
= 30 kN.
Reaction in beam with different end conditions
Questionsfor practice
Question4:FindthereactionsatsupportsAandBofthebeamshowninFigure.
1 m
40 kN
2/2/2023 35
ΣH = 0, gives
H
A
= 0
ΣMA = 0, gives
–20 ×2 ×1 + 60 ×4 + 30 + 20 ×11 –V
B
×9 = 0
∴V
B
= 50 kN.
ΣV = 0, gives
–20 ×2 + V
A
–60 + V
B
–20 = 0
V
A
= 120 –V
B
= 120 –50
V
A
= 70 kN
Questionsfor practice
Question4:FindthereactionsatsupportsAandBofthebeamshowninFigure.
1 m
40 kN
2/2/2023 35
ΣH = 0, gives
H
A
= 0
ΣMA = 0, gives
–20 ×2 ×1 + 60 ×4 + 30 + 20 ×11 –V
B
×9 = 0
∴V
B
= 50 kN.
ΣV = 0, gives
–20 ×2 + V
A
–60 + V
B
–20 = 0
V
A
= 120 –V
B
= 120 –50
V
A
= 70 kN
Reaction in trusses with different end conditions
Trusses
Atrussisastructuremadeupofslendermemberspin-connectedatendsandiscapableoftakingloadsatjoints.
Transmissiontowersarealsotheexamplesoftrusses
PERFECT, DEFICIENT AND REDUNDANT TRUSSES
m = 2j –3
where: m = number of member
J = number of joint
Theaboveequationgivesonlyanecessary,butnotasufficient
conditionofaperfecttruss(Exp.Fig(a)And(b))
m = 2j –3 Perfect Truss
Deficient Truss
Redundant Truss
m 2j –3
m > 2j –3
2/2/2023 36
Trusses
Atrussisastructuremadeupofslendermemberspin-connectedatendsandiscapableoftakingloadsatjoints.
Transmissiontowersarealsotheexamplesoftrusses
PERFECT, DEFICIENT AND REDUNDANT TRUSSES
m = 2j –3
where: m = number of member
J = number of joint
Theaboveequationgivesonlyanecessary,butnotasufficient
conditionofaperfecttruss(Exp.Fig(a)And(b))
m = 2j –3 Perfect Truss
Deficient Truss
Redundant Truss
m 2j –3
m > 2j –3
2/2/2023 36
Reaction in trusses with different end conditions
ASSUMPTIONS
1. The ends of the members are pin-connected (hinged)
2. The loads act only at the joints
3. Self-weights of the members are negligible
4. Cross-section of the members is uniform
NATURE OF FORCES IN MEMBERS
Compressive
Tensile
2/2/2023 37
ASSUMPTIONS
1. The ends of the members are pin-connected (hinged)
2. The loads act only at the joints
3. Self-weights of the members are negligible
4. Cross-section of the members is uniform
NATURE OF FORCES IN MEMBERS
Compressive
Tensile
2/2/2023 37
Reaction in trusses with different end conditions
METHODS OF ANALYSIS
Method of joints Method of section Graphical method
Analytical method
2/2/2023 38
Note: When three members are meeting at an unloaded joint and out of them two are collinear, then the force
in third member will be zero. Such situations are illustrated in Fig.
METHODS OF ANALYSIS
Method of joints Method of section Graphical method
Analytical method
2/2/2023 38
Note: When three members are meeting at an unloaded joint and out of them two are collinear, then the force
in third member will be zero. Such situations are illustrated in Fig.
Reaction in trusses with different end conditions
Question1:DeterminetheforcesinallthemembersofthetrussshowninFig.andindicatethemagnitudeand
natureofforcesonthediagramofthetruss.Allinclinedmembersareat60°tohorizontalandlengthofeach
memberis2m
Solution:
1 m1 m
H
A
= 0
R
D
= 77.5 kN
R
A
= 72.5 kN
2/2/2023 39
Question1:DeterminetheforcesinallthemembersofthetrussshowninFig.andindicatethemagnitudeand
natureofforcesonthediagramofthetruss.Allinclinedmembersareat60°tohorizontalandlengthofeach
memberis2m
Solution:
1 m1 m
H
A
= 0
R
D
= 77.5 kN
R
A
= 72.5 kN
2/2/2023 39
Reaction in trusses with different end conditions
Joint A:
Σ V = 0, gives
F
AB
sin 60°= R
A
= 72.5
F
AB
= 83.7158 kN (Comp.)
Σ H = 0, gives
F
AE
–83.7158 cos 60°= 0
F
AE
= 41.8579 kN (Tension)
Joint D:
Σ V = 0, gives
F
DC
sin 60°= R
D
= 77.5
F
DC
= 89.4893 kN (Comp.)
Σ H = 0, gives
F
DE
–87.4893 cos 60°= 0
F
DE
= 44.7446 kN (Tension)
Joint B:
Σ V = 0, gives
F
BE
sin 60°–F
AB
sin 60°+ 40 = 0
F
BE
=37. 5278(Tension)
Σ H = 0, gives
F
BC
–F
AB
cos 60°–F
BE
cos 60°= 0
F
BC
= (83.7158 + 37.5274) ×0.5
F
BC
= 60.6218 kN (Comp.)
2/2/2023 40
SC2
Joint A:
Σ V = 0, gives
F
AB
sin 60°= R
A
= 72.5
F
AB
= 83.7158 kN (Comp.)
Σ H = 0, gives
F
AE
–83.7158 cos 60°= 0
F
AE
= 41.8579 kN (Tension)
Joint D:
Σ V = 0, gives
F
DC
sin 60°= R
D
= 77.5
F
DC
= 89.4893 kN (Comp.)
Σ H = 0, gives
F
DE
–87.4893 cos 60°= 0
F
DE
= 44.7446 kN (Tension)
Joint B:
Σ V = 0, gives
F
BE
sin 60°–F
AB
sin 60°+ 40 = 0
F
BE
=37. 5278(Tension)
Σ H = 0, gives
F
BC
–F
AB
cos 60°–F
BE
cos 60°= 0
F
BC
= (83.7158 + 37.5274) ×0.5
F
BC
= 60.6218 kN (Comp.)
2/2/2023 40
SC2
Slide 40
SC2
Shailendra Chaurasiya, 1/19/2021
SC2
Shailendra Chaurasiya, 1/19/2021
Reaction in trusses with different end conditions
Joint C:
Σ V = 0, gives
F
CE
sin 60°+ 50 –F
DC
sin 60°= 0
F
CE
=
;;.9 ? 94
??? :4
?
= 31.7543 kN (Tension)
2/2/2023 41
Joint C:
Σ V = 0, gives
F
CE
sin 60°+ 50 –F
DC
sin 60°= 0
F
CE
=
;;.9 ? 94
??? :4
?
= 31.7543 kN (Tension)
2/2/2023 41
Reaction in trusses with different end conditions
Question 2: Determine the forces in the members of truss shown in Fig.
2/2/2023 42
Reaction forces at support
H
A
= 20 kN
R
E
= 90 kN
V
A
=
60 kN
Joint A:
ΣV = 0
F
AB
= 60 kN (Comp.)
ΣH = 0
F
AF
= 20 kN (Tensile)
Joint E:
Σ V = 0
F
ED
= 90 kN (Comp.)
F
EF
= 0
ΣH = 0,
Question 2: Determine the forces in the members of truss shown in Fig.
2/2/2023 42
Reaction forces at support
H
A
= 20 kN
R
E
= 90 kN
V
A
=
60 kN
Joint A:
ΣV = 0
F
AB
= 60 kN (Comp.)
ΣH = 0
F
AF
= 20 kN (Tensile)
Joint E:
Σ V = 0
F
ED
= 90 kN (Comp.)
F
EF
= 0
ΣH = 0,
Reaction in trusses with different end conditions
2/2/2023 43
Joint B:
Σ V = 0
–F
BF
sin 45°–30 + 60 = 0
F
BF
= 30/sin 45
F
BF
= 42.43 kN [Tensile]
Σ H = 0,
–F
BC
+ F
BF
cos 45°= 0
F
BC
= 42.43 ×cos 45°= 30 kN [Comp.]
Joint C:
Σ V = 0, F
CF
= 50 kN [Comp.]
ΣH = 0, F
CD
= 30 kN [Comp.]
Joint D:
ΣV = 0
–F
DF
cos 45°+ 90 –40 = 0
F
DF
= 70.71 kN [Tensile]
2/2/2023 43
Joint B:
Σ V = 0
–F
BF
sin 45°–30 + 60 = 0
F
BF
= 30/sin 45
F
BF
= 42.43 kN [Tensile]
Σ H = 0,
–F
BC
+ F
BF
cos 45°= 0
F
BC
= 42.43 ×cos 45°= 30 kN [Comp.]
Joint C:
Σ V = 0, F
CF
= 50 kN [Comp.]
ΣH = 0, F
CD
= 30 kN [Comp.]
Joint D:
ΣV = 0
–F
DF
cos 45°+ 90 –40 = 0
F
DF
= 70.71 kN [Tensile]
Reaction in trusses with different end conditions
Question 3: Analyze the truss shown in Fig.
2/2/2023 44
Joint E:
F
ED
×Sin θ-20 = 0
F
ED
= 25 kN (Tension)
ΣH = 0,
F
EF
–F
ED
cos θ= 0
F
EF
= 25 ×cos θ= 15 kN (Comp.)
ΣM
A
= 0,
Reaction at Support
R
C
= 15 kN
ΣV = 0
V
A
= 20 kN
ΣH = 0,
H
A
= R
C
= 15 kN
Joint A:
F
AB
–V
A
= 0
F
AB
= 20 kN (Comp.)
F
AF
–H
A
= 0
F
AF
= 15 kN (Comp.)
ΣV = 0, gives
ΣH = 0, gives
Joint C:
ΣV = 0, gives
ΣH = 0, gives
F
CB
×cos θ–R
C
= 0
F
CB
=25 kN (Comp.)
F
CD
= F
CB
sin θ
= 20 kN (Tension)
ΣV = 0, gives
Question 3: Analyze the truss shown in Fig.
2/2/2023 44
Joint E:
F
ED
×Sin θ-20 = 0
F
ED
= 25 kN (Tension)
ΣH = 0,
F
EF
–F
ED
cos θ= 0
F
EF
= 25 ×cos θ= 15 kN (Comp.)
ΣM
A
= 0,
Reaction at Support
R
C
= 15 kN
ΣV = 0
V
A
= 20 kN
ΣH = 0,
H
A
= R
C
= 15 kN
Joint A:
F
AB
–V
A
= 0
F
AB
= 20 kN (Comp.)
F
AF
–H
A
= 0
F
AF
= 15 kN (Comp.)
ΣV = 0, gives
ΣH = 0, gives
Joint C:
ΣV = 0, gives
ΣH = 0, gives
F
CB
×cos θ–R
C
= 0
F
CB
=25 kN (Comp.)
F
CD
= F
CB
sin θ
= 20 kN (Tension)
ΣV = 0, gives
Reaction in trusses with different end conditions
Joint B:
ΣV = 0, gives
F
BF
×sin θ–F
BC
×sin θ+ F
AB
= 0
F
BF
= 0
Σ H = 0, gives
F
BD
–25 ×cos θ= 0
F
BD
= 15 kN (Tension)
Joint F:
ΣV = 0
F
FD
= 0
2/2/2023 45
Joint B:
ΣV = 0, gives
F
BF
×sin θ–F
BC
×sin θ+ F
AB
= 0
F
BF
= 0
Σ H = 0, gives
F
BD
–25 ×cos θ= 0
F
BD
= 15 kN (Tension)
Joint F:
ΣV = 0
F
FD
= 0
2/2/2023 45
Reaction in trusses with different end conditions
Question for Practice
Question 1: Determine the forces in the members of truss shown in Fig. All the members are of 3 m length
Question 2: Determine the forces in the members of truss shown in Fig.
2/2/2023 46
Question for Practice
Question 1: Determine the forces in the members of truss shown in Fig. All the members are of 3 m length
Question 2: Determine the forces in the members of truss shown in Fig.
2/2/2023 46
Reaction in trusses with different end conditions
METHOD OF SECTION
Question1:DeterminetheforcesinthemembersFH,HGandGIinthetrussshowninFig.Eachloadis10kNandall
trianglesareequilateralwithsides4m.
R
A
= R
0
=
5
6
×10 ×7 = 35 kN = R
B
Left Section
Reaction force at support
ΣM
G
= 0,
F
FH
×4 sin 60°–35 ×12 + 10 ×10 + 10 ×6 + 10 ×2 = 0
F
FH
= 69.2820 kN (Comp.)
ΣV = 0, gives
F
GH
sin 60°+ 10 + 10 + 10 –35 = 0
F
GH
= 5.7735 kN (Comp.)
Σ H = 0, gives
F
GI
–F
FH
–F
GH
cos 60°= 0
F
GI
= 69.2820 + 5.7735 cos 60°= 72.1688 kN (Tension)
Left Section Right Section
2/2/2023 47
METHOD OF SECTION
Question1:DeterminetheforcesinthemembersFH,HGandGIinthetrussshowninFig.Eachloadis10kNandall
trianglesareequilateralwithsides4m.
R
A
= R
0
=
5
6
×10 ×7 = 35 kN = R
B
Left Section
Reaction force at support
ΣM
G
= 0,
F
FH
×4 sin 60°–35 ×12 + 10 ×10 + 10 ×6 + 10 ×2 = 0
F
FH
= 69.2820 kN (Comp.)
ΣV = 0, gives
F
GH
sin 60°+ 10 + 10 + 10 –35 = 0
F
GH
= 5.7735 kN (Comp.)
Σ H = 0, gives
F
GI
–F
FH
–F
GH
cos 60°= 0
F
GI
= 69.2820 + 5.7735 cos 60°= 72.1688 kN (Tension)
Left Section Right Section
2/2/2023 47
Reaction in trusses with different end conditions
Question2:
FindthemagnitudeandnatureoftheforcesinthemembersU3U4,L3L4andU4L3oftheloadedtrussshowninFig.
Left Section Right Section
ΣM
LO
= 0, gives
R
2
×36 –200 ×6 –200 ×12 –150
×18 –100 ×24 –100 ×30 = 0
R
2
= 325 kN
ΣV = 0, gives
R
1
= 200 + 200 + 150 + 100 + 100 –325
R
1
= 425 kN
U
3
U
4
= 1
6
+6
6
Sin θ
1
=
5
6.0828
= 0.1644
Cos θ
1
=
:
6.0828
= 0.9864
L
3
U
4
= 8
6
+6
6
= 10
sin θ2 = 0.6 cos θ2 = 0.8
F
L3L4
×8 –325 ×12 + 100 ×6 = 0
F
L3L4
= 412.5 kN (Tension)
ΣM
L3
= 0, gives
F
U4U3
×cos θ
1
×9 + 100 ×6 + 100 ×12 –325 ×18 = 0
F
U4U3
= 456.2072 kN (Comp.)
Σ H = 0
F
U4U3
sin θ
2
–F
U4U3
cos θ
1
+ F
L4L3
= 0
F
U4U3
= 62.5 kN (Tension)
Reaction at Support
Right Section
2/2/2023 48
Question2:
FindthemagnitudeandnatureoftheforcesinthemembersU3U4,L3L4andU4L3oftheloadedtrussshowninFig.
Left Section Right Section
ΣM
LO
= 0, gives
R
2
×36 –200 ×6 –200 ×12 –150
×18 –100 ×24 –100 ×30 = 0
R
2
= 325 kN
ΣV = 0, gives
R
1
= 200 + 200 + 150 + 100 + 100 –325
R
1
= 425 kN
U
3
U
4
= 1
6
+6
6
Sin θ
1
=
5
6.0828
= 0.1644
Cos θ
1
=
:
6.0828
= 0.9864
L
3
U
4
= 8
6
+6
6
= 10
sin θ2 = 0.6 cos θ2 = 0.8
F
L3L4
×8 –325 ×12 + 100 ×6 = 0
F
L3L4
= 412.5 kN (Tension)
ΣM
L3
= 0, gives
F
U4U3
×cos θ
1
×9 + 100 ×6 + 100 ×12 –325 ×18 = 0
F
U4U3
= 456.2072 kN (Comp.)
Σ H = 0
F
U4U3
sin θ
2
–F
U4U3
cos θ
1
+ F
L4L3
= 0
F
U4U3
= 62.5 kN (Tension)
Reaction at Support
Right Section
2/2/2023 48
Reaction in trusses with different end conditions
Question for Practice
Ans.
F
AC
= –67.48 kN;
F
AB
= +53.99 kN;
F
BC
= +10 kN;
F
CD
= –8.33 kN;
F
CE
= –59.15 kN;
F
EF
= –24.5 kN;
F
ED
= +52.81 kN;
F
FD
= +47.21 kN;
F
FG
= –34.64 kN;
F
DG
= +47.32 kN
Ans.
F
AB
= +10 13kN;
F
AC
= –20 kN;
F
CB
= –48.75 kN;
F
CE
= –20 kN;
F
CD
= –7.5 kN;
F
BE
= +6.25 13kN;
F
DE
= 18.75 kN;
F
DF
= -3.75 13kN;
F
FE
= –7.5 kN
Question 1: Question 2:
2/2/2023 49
Question for Practice
Ans.
F
AC
= –67.48 kN;
F
AB
= +53.99 kN;
F
BC
= +10 kN;
F
CD
= –8.33 kN;
F
CE
= –59.15 kN;
F
EF
= –24.5 kN;
F
ED
= +52.81 kN;
F
FD
= +47.21 kN;
F
FG
= –34.64 kN;
F
DG
= +47.32 kN
Ans.
F
AB
= +10 13kN;
F
AC
= –20 kN;
F
CB
= –48.75 kN;
F
CE
= –20 kN;
F
CD
= –7.5 kN;
F
BE
= +6.25 13kN;
F
DE
= 18.75 kN;
F
DF
= -3.75 13kN;
F
FE
= –7.5 kN
Question 1: Question 2:
2/2/2023 49
Centre of gravity and moment of inertia
Area of Standard Figures
1. Area of a rectangle
Area of rectangle, A = bd
2. Area of a triangle
Area of tangle, A =
bh
2
3.Area of a circle
Area of Cirle, A = πR
2
4. Area of a sector of a circle
5. Surface area of a sphere
Area of sphere, A =4πR
2
2/2/2023 50
Area of sector of a circle, A =R
2
α
Area of Standard Figures
1. Area of a rectangle
Area of rectangle, A = bd
2. Area of a triangle
Area of tangle, A =
bh
2
3.Area of a circle
Area of Cirle, A = πR
2
4. Area of a sector of a circle
5. Surface area of a sphere
Area of sphere, A =4πR
2
2/2/2023 50
Area of sector of a circle, A =R
2
α
Centre of gravity and moment of inertia
2/2/2023 51
2/2/2023 51
Centre of gravity and moment of inertia
Centroid of Some Common Figures
FIRST MOMENT OF AREA AND CENTROID
x
c
=
∑
A
i
xi
A
y
c
=
∑
A
i
y
i
A
CENTRE OF GRAVITY AND CENTROIDS
Centreofgravityisthepointatwhichtheentire
weightofabodymaybeconsideredas
concentratedsothatifsupportedatthispointthe
bodywouldremaininequilibriuminanyposition .
2/2/2023 52
Centroidisthepointthatmaybeconsideredasthe
centerofaone-ortwo-dimensionalfigure,thesum
ofthedisplacementsofallpointsinthefigurefrom
suchapointbeingzero
Centroid of Some Common Figures
FIRST MOMENT OF AREA AND CENTROID
x
c
=
∑
A
i
xi
A
y
c
=
∑
A
i
y
i
A
CENTRE OF GRAVITY AND CENTROIDS
Centreofgravityisthepointatwhichtheentire
weightofabodymaybeconsideredas
concentratedsothatifsupportedatthispointthe
bodywouldremaininequilibriuminanyposition .
2/2/2023 52
Centroidisthepointthatmaybeconsideredasthe
centerofaone-ortwo-dimensionalfigure,thesum
ofthedisplacementsofallpointsinthefigurefrom
suchapointbeingzero
Centre of gravity and moment of inertia
Question 1. Locate the centroid of the T-section shown in the Fig.
A
1
A
2
A
1
= 20 ×100 = 2000
x
1
= 0
Y
1
= 10
Centroid T -Section
??????̅=
A
1
× x
1
+ A
2
× x
2
A
1
+ A
2
=
2000 × 0 + 2000 × 0
2000 + 2000
??????̅= 0
A
2
= 20 ×100 = 2000
x
2
= 0
Y
2
= 70
?????? %=
A
1
× y
1
+ A
2
× y
2
A
1
+ A
2
=
2000 × 10 + 2000 × 70
2000 + 2000
?????? % = 40
Section 1
Section 2
2/2/2023 53
Question 1. Locate the centroid of the T-section shown in the Fig.
A
1
A
2
A
1
= 20 ×100 = 2000
x
1
= 0
Y
1
= 10
Centroid T -Section
??????̅=
A
1
× x
1
+ A
2
× x
2
A
1
+ A
2
=
2000 × 0 + 2000 × 0
2000 + 2000
??????̅= 0
A
2
= 20 ×100 = 2000
x
2
= 0
Y
2
= 70
?????? %=
A
1
× y
1
+ A
2
× y
2
A
1
+ A
2
=
2000 × 10 + 2000 × 70
2000 + 2000
?????? % = 40
Section 1
Section 2
2/2/2023 53
Centre of gravity and moment of inertia
Question 2. Locate the centroid of the I-section shown in Fig.
A
1
A
2
A
3
A
1
= 20 ×100 = 2000 mm
2
x
1
= 0
Y
1
= 90
Centroid T -Section
??????̅=
A
1
× x
1
+ A
2
× x
2
+ A
3
× x
3
A
1
+A
2
+ A
3
=
2000 × 0 + 2000 × 0 + 4500 × 0
2000 + 2000 + 4500
??????%= 0 mm
A
2
= 20 ×100 = 2000 mm
2
x
2
= 0
Y
2
= 80
?????? %=
A
1
× y
1
+ A
2
× y
2
+ A
3
× y
3
A
1
+ A
2
+ A
7
=
200 × 90 + 200 × 80 + 4500 × 70
2000 + 2000+ 4500
?????? % = 59.71 mm
Section 1 Section 2
Section 3
A
2
= 30 ×150 = 4500 mm
2
x
2
= 0
Y
2
= 70
2/2/2023 54
Question 2. Locate the centroid of the I-section shown in Fig.
A
1
A
2
A
3
A
1
= 20 ×100 = 2000 mm
2
x
1
= 0
Y
1
= 90
Centroid T -Section
??????̅=
A
1
× x
1
+ A
2
× x
2
+ A
3
× x
3
A
1
+A
2
+ A
3
=
2000 × 0 + 2000 × 0 + 4500 × 0
2000 + 2000 + 4500
??????%= 0 mm
A
2
= 20 ×100 = 2000 mm
2
x
2
= 0
Y
2
= 80
?????? %=
A
1
× y
1
+ A
2
× y
2
+ A
3
× y
3
A
1
+ A
2
+ A
7
=
200 × 90 + 200 × 80 + 4500 × 70
2000 + 2000+ 4500
?????? % = 59.71 mm
Section 1 Section 2
Section 3
A
2
= 30 ×150 = 4500 mm
2
x
2
= 0
Y
2
= 70
2/2/2023 54
Centre of gravity and moment of inertia
Question 2. Locate the centroid of the section shown in Fig.
A
1
=
5
6
×3 ×4 = 6 m
2
x
1
= 6 + 1 = 7 m
Y
1
=
8
7
=1.333 m
Centroid T -Section
??????̅=
A
1
× x
1
+ A
2
× x
2
+ A
3
× x
3
A
1
+A
2
+ A
3
=
6 × 7 + 24 × 3 + 6.2831 ×
? 4.<8<<
6 + 24 +
:.6<75
??????%= 2.99 m
A
2
= 4 ×6 = 24 m
2
x
2
= 3 m
Y
2
= 2 m
?????? %=
A
1
× y
1
+ A
2
× y
2
+ A
3
× y
3
A
1
+ A
2
+ A
7
=
6 × 1.333 + 24 × 2 + 6.2831 × 2
6 + 24+ 6.2831
?????? % = 1.8892 m
Section 1 Section 2
Section 3
A
2
=
5
6
×π ×2
2
= 6.2831 m
2
x
2
= –
8?
7
π
= –0.8488 m
Y
2
= 2 m
A
1
A
2
A
3
2/2/2023 55
Question 2. Locate the centroid of the section shown in Fig.
A
1
=
5
6
×3 ×4 = 6 m
2
x
1
= 6 + 1 = 7 m
Y
1
=
8
7
=1.333 m
Centroid T -Section
??????̅=
A
1
× x
1
+ A
2
× x
2
+ A
3
× x
3
A
1
+A
2
+ A
3
=
6 × 7 + 24 × 3 + 6.2831 ×
? 4.<8<<
6 + 24 +
:.6<75
??????%= 2.99 m
A
2
= 4 ×6 = 24 m
2
x
2
= 3 m
Y
2
= 2 m
?????? %=
A
1
× y
1
+ A
2
× y
2
+ A
3
× y
3
A
1
+ A
2
+ A
7
=
6 × 1.333 + 24 × 2 + 6.2831 × 2
6 + 24+ 6.2831
?????? % = 1.8892 m
Section 1 Section 2
Section 3
A
2
=
5
6
×π ×2
2
= 6.2831 m
2
x
2
= –
8?
7
π
= –0.8488 m
Y
2
= 2 m
A
1
A
2
A
3
2/2/2023 55
Centre of gravity and moment of inertia
Question 4. Determine the centroid of the section of the concrete dam shown in Fig
Centroid of Section
The centroid x = 3.523 m
and y = 2.777 m
2/2/2023 56
Question 4. Determine the centroid of the section of the concrete dam shown in Fig
Centroid of Section
The centroid x = 3.523 m
and y = 2.777 m
2/2/2023 56
Centre of gravity and moment of inertia
Question 5. Determine the coordinates x
c
and y
c
of the center of a 100 mm diameter circular hole cut in a thin plate so that
this point will be the centroid of the remaining shaded area shown in Fig. (All dimensions are in mm).
x
c
(18396) = 200 ×150 ×100 -
5
6
×100 ×75 ×166.67 –4×100
2
x
c
18396 y
c
= 200 ×150 ×75 –
5
6
×100 ×75 ×(150 –25) –
8
×100
2
y
c
2/2/2023 57
Question 5. Determine the coordinates x
c
and y
c
of the center of a 100 mm diameter circular hole cut in a thin plate so that
this point will be the centroid of the remaining shaded area shown in Fig. (All dimensions are in mm).
x
c
(18396) = 200 ×150 ×100 -
5
6
×100 ×75 ×166.67 –4×100
2
x
c
18396 y
c
= 200 ×150 ×75 –
5
6
×100 ×75 ×(150 –25) –
8
×100
2
y
c
2/2/2023 57
Centre of gravity and moment of inertia
Question 6: Determine the coordinates of the centroid of the plane area shown in Fig. with reference to the axis shown.
Take x = 40 mm.
= 8.1599x
= 8.1599 ×40
= 326.40 mm
2/2/2023 58
Question 6: Determine the coordinates of the centroid of the plane area shown in Fig. with reference to the axis shown.
Take x = 40 mm.
= 8.1599x
= 8.1599 ×40
= 326.40 mm
2/2/2023 58
Centre of gravity and moment of inertia
Question 7: Determine the centroid of the wire shown in Fig.
2/2/2023 59
Question 7: Determine the centroid of the wire shown in Fig.
2/2/2023 59
Centre of gravity and moment of inertia
Moments of Inertia
•Themoment of inertiais a measure of how resistant an object is to changes in its rotational motion.
•Themoment of inertiaof the cross-section of a body is its resistance to changes in its rotation.
•It depends on how far each part of the body's mass is from its center.
Theorems of Moments of Inertia
1.Perpendicular axis theorem
Themomentofinertiaofanareaaboutanaxis
perpendiculartoitsplane(polarmomentofinertia)
atanypointOisequaltothesumofmomentsof
inertiaaboutanytwomutuallyperpendicularaxis
throughthesamepointOandlyingintheplaneof
thearea.
2. Parallel axis theorem
Momentofinertiaaboutanyaxisintheplaneofan
areaisequaltothesumofmomentofinertiaabouta
parallelcentroidalaxisandtheproductofareaand
squareofthemdistancebetweenthetwoparallel
axis
I
zz
=r
2
dA
Izz = Ixx + Iyy
2/2/2023 60
Moments of Inertia
•Themoment of inertiais a measure of how resistant an object is to changes in its rotational motion.
•Themoment of inertiaof the cross-section of a body is its resistance to changes in its rotation.
•It depends on how far each part of the body's mass is from its center.
Theorems of Moments of Inertia
1.Perpendicular axis theorem
Themomentofinertiaofanareaaboutanaxis
perpendiculartoitsplane(polarmomentofinertia)
atanypointOisequaltothesumofmomentsof
inertiaaboutanytwomutuallyperpendicularaxis
throughthesamepointOandlyingintheplaneof
thearea.
2. Parallel axis theorem
Momentofinertiaaboutanyaxisintheplaneofan
areaisequaltothesumofmomentofinertiaabouta
parallelcentroidalaxisandtheproductofareaand
squareofthemdistancebetweenthetwoparallel
axis
I
zz
=r
2
dA
Izz = Ixx + Iyy
2/2/2023 60
Centre of gravity and moment of inertia
Moment of Inertia of Standard Sections
2/2/2023 61
Moment of Inertia of Standard Sections
2/2/2023 61
Centre of gravity and moment of inertia
Moment of Inertia of Standard Sections
2/2/2023 62
Moment of Inertia of Standard Sections
2/2/2023 62
Centre of gravity and moment of inertia
Moment of Inertia of Standard Sections
2/2/2023 63
Moment of Inertia of Standard Sections
2/2/2023 63
Centre of gravity and moment of inertia
Question1.DeterminethemomentofinertiaofthesectionshowninFig.aboutanaxispassingthroughthecentroidand
paralleltothetopmostfibreofthesection.Alsodeterminemomentofinertiaabouttheaxisofsymmetry.Hencefindradii
ofgyration.
A
1
= 150 ×10 = 1500 mm
2
A
2
= 140 ×10 = 1400 mm
2
A = A
1
+ A
2
= 2900 mm
2
Centroid (x
c
, y
c
) = (0.0, 41.21)
Moment of inertia of the section about x-x axis
I
xx
= moment of inertia of A
1
about x-x axis + moment of inertia of A
2
about x-x axis
I
xx
= {
??
7
56
+ A
1
×(Y
1
-Y
c
)
2
}
1
+
{
??
7
56
+ A
2
×(
Y
2
-Y
c
)
2
}
2
I
yy
= {
??
7
56
+ A
1
×(x
1
-x
c
)
2
}
1
+
{
??
7
56
+ A
2
×(
x
2
-x
c
)
2
}
2
I
xx
= 63 72442.5 mm
4
I
yy
= 2824,166.7 mm
4
Moment of inertia of the section about y-y axis
I
yy
= moment of inertia of A
1
about y-y axis + moment of inertia of A
2
about y-y axis
2/2/2023 64
Question1.DeterminethemomentofinertiaofthesectionshowninFig.aboutanaxispassingthroughthecentroidand
paralleltothetopmostfibreofthesection.Alsodeterminemomentofinertiaabouttheaxisofsymmetry.Hencefindradii
ofgyration.
A
1
= 150 ×10 = 1500 mm
2
A
2
= 140 ×10 = 1400 mm
2
A = A
1
+ A
2
= 2900 mm
2
Centroid (x
c
, y
c
) = (0.0, 41.21)
Moment of inertia of the section about x-x axis
I
xx
= moment of inertia of A
1
about x-x axis + moment of inertia of A
2
about x-x axis
I
xx
= {
??
7
56
+ A
1
×(Y
1
-Y
c
)
2
}
1
+
{
??
7
56
+ A
2
×(
Y
2
-Y
c
)
2
}
2
I
yy
= {
??
7
56
+ A
1
×(x
1
-x
c
)
2
}
1
+
{
??
7
56
+ A
2
×(
x
2
-x
c
)
2
}
2
I
xx
= 63 72442.5 mm
4
I
yy
= 2824,166.7 mm
4
Moment of inertia of the section about y-y axis
I
yy
= moment of inertia of A
1
about y-y axis + moment of inertia of A
2
about y-y axis
2/2/2023 64
Centre of gravity and moment of inertia
The radius of gyration is given by:
I
xx
= 63 72442.5 mm
4
I
yy
= 2824,166.7 mm
4
k
xx
= 46.88 mm
k
yy
= 31.21 mm
Similarly,
2/2/2023 65
The radius of gyration is given by:
I
xx
= 63 72442.5 mm
4
I
yy
= 2824,166.7 mm
4
k
xx
= 46.88 mm
k
yy
= 31.21 mm
Similarly,
2/2/2023 65
Centre of gravity and moment of inertia
Question 2.Determine the moment of inertia of the L-section shown in the Fig. about its centroidal axis parallel to the
legs. Also find out the polar moment of inertia.
A
1
= 125 ×10 = 1250 mm
2
A
2
= 75 ×10 = 750 mm
2
Total Area (A) = 2000 mm
2
Centroid (x
c
, y
c
) = (20.94, 40.94)
Moment of inertia of the section about x-x axis
I
xx
= moment of inertia of A
1
about x-x axis + moment of inertia of A
2
about x-x axis
I
xx
= {
??
7
56
+ A
1
×(Y
1
-Y
c
)
2
}
1
+
{
??
7
56
+ A
2
×(
Y
2
-Y
c
)
2
}
2
I
yy
= {
??
7
56
+ A
1
×(x
1
-x
c
)
2
}
1
+
{
??
7
56
+ A
2
×(
x
2
-x
c
)
2
}
2
I
xx
= 3411298.9 mm
4
I
yy
= 1208658.9mm
4
Moment of inertia of the section about y-y axis
I
yy
= moment of inertia of A
1
about y-y axis + moment of inertia of A
2
about y-y axis
2/2/2023 66
Question 2.Determine the moment of inertia of the L-section shown in the Fig. about its centroidal axis parallel to the
legs. Also find out the polar moment of inertia.
A
1
= 125 ×10 = 1250 mm
2
A
2
= 75 ×10 = 750 mm
2
Total Area (A) = 2000 mm
2
Centroid (x
c
, y
c
) = (20.94, 40.94)
Moment of inertia of the section about x-x axis
I
xx
= moment of inertia of A
1
about x-x axis + moment of inertia of A
2
about x-x axis
I
xx
= {
??
7
56
+ A
1
×(Y
1
-Y
c
)
2
}
1
+
{
??
7
56
+ A
2
×(
Y
2
-Y
c
)
2
}
2
I
yy
= {
??
7
56
+ A
1
×(x
1
-x
c
)
2
}
1
+
{
??
7
56
+ A
2
×(
x
2
-x
c
)
2
}
2
I
xx
= 3411298.9 mm
4
I
yy
= 1208658.9mm
4
Moment of inertia of the section about y-y axis
I
yy
= moment of inertia of A
1
about y-y axis + moment of inertia of A
2
about y-y axis
2/2/2023 66
Centre of gravity and moment of inertia
Question 2.Determine the moment of inertia of the L-section shown in the Fig. about its centroidal axis parallel to the
legs. Also find out the polar moment of inertia.
Polar moment of inertia (I
zz
)
(I
zz
) = I
xx
+ I
yy
= 3411298.9 + 12,08658.9
I
zz
= 4619957.8 mm
4
2/2/2023 67
Question 2.Determine the moment of inertia of the L-section shown in the Fig. about its centroidal axis parallel to the
legs. Also find out the polar moment of inertia.
Polar moment of inertia (I
zz
)
(I
zz
) = I
xx
+ I
yy
= 3411298.9 + 12,08658.9
I
zz
= 4619957.8 mm
4
2/2/2023 67
Centre of gravity and moment of inertia
Question 3. Determine the moment of inertia of the symmertic I-section shown in Fig. about its centroidal axis x-x and y-y.
A
1
= -mm
2
A
2
= -mm
2
A
3
= -mm
2
Total Area (A) = ----mm
2
Centroid (x
c
, y
c
) = (-, -)
Moment of inertia of the section about x-x axis
I
xx
= moment of inertia of A
1
about x-x axis + moment of inertia of A
2
about x-x axis
I
xx
= {
??
7
56
+ A
1
×(Y
1
-Y
c
)
2
}
1
+
{
??
7
56
+ A
2
×(
Y
2
-Y
c
)
2
}
2
+{
??
7
56
+ A
3
×(
Y
3
-Y
c
)
2
}
3
I
yy
= {
??
7
56
+ A
1
×(x
1
-x
c
)
2
}
1
+
{
??
7
56
+ A
2
×(
x
2
-x
c
)
2
}
2
+{
??
7
56
+ A
3
×(
x
3
-x
c
)
2
}
3
I
xx
= 5,92,69,202 mm
4
I
yy
= 1,20,05,815 m
4
Moment of inertia of the section about y-y axis
I
yy
= moment of inertia of A
1
about y-y axis + moment of inertia of A
2
about y-y axis
2/2/2023 68
Question 3. Determine the moment of inertia of the symmertic I-section shown in Fig. about its centroidal axis x-x and y-y.
A
1
= -mm
2
A
2
= -mm
2
A
3
= -mm
2
Total Area (A) = ----mm
2
Centroid (x
c
, y
c
) = (-, -)
Moment of inertia of the section about x-x axis
I
xx
= moment of inertia of A
1
about x-x axis + moment of inertia of A
2
about x-x axis
I
xx
= {
??
7
56
+ A
1
×(Y
1
-Y
c
)
2
}
1
+
{
??
7
56
+ A
2
×(
Y
2
-Y
c
)
2
}
2
+{
??
7
56
+ A
3
×(
Y
3
-Y
c
)
2
}
3
I
yy
= {
??
7
56
+ A
1
×(x
1
-x
c
)
2
}
1
+
{
??
7
56
+ A
2
×(
x
2
-x
c
)
2
}
2
+{
??
7
56
+ A
3
×(
x
3
-x
c
)
2
}
3
I
xx
= 5,92,69,202 mm
4
I
yy
= 1,20,05,815 m
4
Moment of inertia of the section about y-y axis
I
yy
= moment of inertia of A
1
about y-y axis + moment of inertia of A
2
about y-y axis
2/2/2023 68
Centre of gravity and moment of inertia
Question 4: Compute the second moment of area of the channel section shown in Fig. about centroidal axis x-x and y-y.
A1 = -mm
2
A2 = -mm
2
A
3
= -mm
2
Total Area (A) = ----mm
2
Centroid (x
c
, y
c
) = (25.73, 0)
Moment of inertia of the section about x-x axis
I
xx
= moment of inertia of A
1
about x-x axis + moment of inertia of A
2
about x-x axis
I
xx
= {
??
7
56
2
+ A
1
×(Y
1
-Y
c
)
2
}
1
+
{
??
7
56
2
+ A
2
×(
Y
2
-Y
c
)
2
}
2
+{
??
7
56
2
+ A
3
×(
Y
3
-Y
c
)
2
}
3
I
yy
= {
??
7
56
2
+ A
1
×(x
1
-x
c
)
2
}
1
+
{
??
7
56
2
+ A
2
×(
x
2
-x
c
)
2
}
2
+{
??
7
56
2
+ A
3
×(
x
3
-x
c
)
2
}
3
I
xx
=1.359 ×10
8
m
4
I
yy
= 52,72557.6 mm
4
Moment of inertia of the section about y-y axis
I
yy
= moment of inertia of A
1
about y-y axis + moment of inertia of A
2
about y-y axis
, 0 69
Question 4: Compute the second moment of area of the channel section shown in Fig. about centroidal axis x-x and y-y.
A1 = -mm
2
A2 = -mm
2
A
3
= -mm
2
Total Area (A) = ----mm
2
Centroid (x
c
, y
c
) = (25.73, 0)
Moment of inertia of the section about x-x axis
I
xx
= moment of inertia of A
1
about x-x axis + moment of inertia of A
2
about x-x axis
I
xx
= {
??
7
56
2
+ A
1
×(Y
1
-Y
c
)
2
}
1
+
{
??
7
56
2
+ A
2
×(
Y
2
-Y
c
)
2
}
2
+{
??
7
56
2
+ A
3
×(
Y
3
-Y
c
)
2
}
3
I
yy
= {
??
7
56
2
+ A
1
×(x
1
-x
c
)
2
}
1
+
{
??
7
56
2
+ A
2
×(
x
2
-x
c
)
2
}
2
+{
??
7
56
2
+ A
3
×(
x
3
-x
c
)
2
}
3
I
xx
=1.359 ×10
8
m
4
I
yy
= 52,72557.6 mm
4
Moment of inertia of the section about y-y axis
I
yy
= moment of inertia of A
1
about y-y axis + moment of inertia of A
2
about y-y axis
, 0 69
Centre of gravity and moment of inertia
Question 5:Determine the moment of inertia of the built-up section shown in Fig. about its centroidal axis x-x and y-y.
Centroid (x
c
, y
c
) = (0, 59.26)
Moment of inertia of the section about x-x axis
I
xx
= moment of inertia of A
1
about x-x axis + moment of inertia of A
2
about x-x axis
I
xx
= {
??????????????????
+ A
i
×(Y
i
-Y
c
)
2
}
i
I
yy
= {
??????????????????
+ A
i
×(x
i
-x
c
)
2
}
i
I
yy
= 1,97,45,122 mm
4
Moment of inertia of the section about y-y axis
I
yy
= moment of inertia of A
1
about y-y axis + moment of inertia of A
2
about y-y axisA
1
= -mm
2
A
2
= -mm
2
A
3
= -mm
2
A
4
= -mm
2
A
5
= -mm
2
Total Area (A) = ----mm
2
I
xx
= 3,15,43,447 mm
4
2/2/2023 70
Question 5:Determine the moment of inertia of the built-up section shown in Fig. about its centroidal axis x-x and y-y.
Centroid (x
c
, y
c
) = (0, 59.26)
Moment of inertia of the section about x-x axis
I
xx
= moment of inertia of A
1
about x-x axis + moment of inertia of A
2
about x-x axis
I
xx
= {
??????????????????
+ A
i
×(Y
i
-Y
c
)
2
}
i
I
yy
= {
??????????????????
+ A
i
×(x
i
-x
c
)
2
}
i
I
yy
= 1,97,45,122 mm
4
Moment of inertia of the section about y-y axis
I
yy
= moment of inertia of A
1
about y-y axis + moment of inertia of A
2
about y-y axisA
1
= -mm
2
A
2
= -mm
2
A
3
= -mm
2
A
4
= -mm
2
A
5
= -mm
2
Total Area (A) = ----mm
2
I
xx
= 3,15,43,447 mm
4
2/2/2023 70
Centre of gravity and moment of inertia
Question:6 Find the second moment of the shaded portion shown in the Fig. about its centroidal axis.
Total Area = Area of triangle ABC + Area of
rectangle ACDE –Area of semicircle
A = 3371.68
Centroid (x
c
, y
c
) = (39.21, 28.47)
Moment of inertia of the section about x-x axis
I
xx
=momentofinertiaofTriangleABC
aboutx-xaxis+momentofinertiaof
RectangleACDEaboutx-xaxis-moment
ofinertiaofsemicircleaboutx-xaxis
I
xx
= {
??????????????????
+A
i
×(Y
i
-Y
c
)
2
}
i
I
yy
=momentofinertiaofTriangleABCabouty-yaxis+momentofinertia
ofRectangleACDEabouty-yaxis-momentofinertiaofsemicircleabout
y-yaxis
Moment of inertia of the section about y-y axis
I
xx
= 6,86,944 mm
4
I
yy
= {
??????????????????
+ A
i
×(x
i
-x
c
)
2
}
i
I
yy
= 1868392 mm
4
2/2/2023 71
Question:6 Find the second moment of the shaded portion shown in the Fig. about its centroidal axis.
Total Area = Area of triangle ABC + Area of
rectangle ACDE –Area of semicircle
A = 3371.68
Centroid (x
c
, y
c
) = (39.21, 28.47)
Moment of inertia of the section about x-x axis
I
xx
=momentofinertiaofTriangleABC
aboutx-xaxis+momentofinertiaof
RectangleACDEaboutx-xaxis-moment
ofinertiaofsemicircleaboutx-xaxis
I
xx
= {
??????????????????
+A
i
×(Y
i
-Y
c
)
2
}
i
I
yy
=momentofinertiaofTriangleABCabouty-yaxis+momentofinertia
ofRectangleACDEabouty-yaxis-momentofinertiaofsemicircleabout
y-yaxis
Moment of inertia of the section about y-y axis
I
xx
= 6,86,944 mm
4
I
yy
= {
??????????????????
+ A
i
×(x
i
-x
c
)
2
}
i
I
yy
= 1868392 mm
4
2/2/2023 71
Stress and Strain
2/2/2023 72
2/2/2023 72
Stress and strain
STRESS (p):
•Internalresistingforcesperunitcross-sectionalarea.
•Theresistingforcenormaltotheplaneiscalled
normalresistance.
•Theresistingforceparalleltotheplaneiscalled
shearingresistance.
Stress =
???????? ????????? ?????
????????? ????
STRAIN (e):
FACTOR OF SAFETY
•Maximum stress to which any member is designed is much less
than the ultimate stress
•The ratio of ultimate stress to working stress is called factor of
safety.
HOOKE’S LAW
•Stress is proportional to strain up to elastic limit
p ∝e
p = Ee
Where: E = modulus of elasticity or Young’s modulus
EXTENSION/SHORTENING OF A BAR
2/2/2023 73
STRESS (p):
•Internalresistingforcesperunitcross-sectionalarea.
•Theresistingforcenormaltotheplaneiscalled
normalresistance.
•Theresistingforceparalleltotheplaneiscalled
shearingresistance.
Stress =
???????? ????????? ?????
????????? ????
STRAIN (e):
FACTOR OF SAFETY
•Maximum stress to which any member is designed is much less
than the ultimate stress
•The ratio of ultimate stress to working stress is called factor of
safety.
HOOKE’S LAW
•Stress is proportional to strain up to elastic limit
p ∝e
p = Ee
Where: E = modulus of elasticity or Young’s modulus
EXTENSION/SHORTENING OF A BAR
2/2/2023 73
Stress and strain
Question 1: A circular rod of diameter 16 mm and 500 mm long is subjected to a tensile force 40 kN. The modulus of
elasticity for steel may be taken as 200 kN/mm
2
. Find stress, strain and elongation of the bar due to applied load
Given that:
Load P = 40 kN = 40 ×1000 N
E = 200 kN/mm
2
= 200 ×10
3
N/mm
2
L = 500 mm
Diameter of the rod d = 16 mm
To find :
•Stress
•Stain
•Elongation in the bar
A = 201.06 mm
2
Stress
p = 198.94 N/mm
2
Strain
e = 0.0009947
Elongation in the bar
∆= 0.497 mm
2/2/2023 74
Question 1: A circular rod of diameter 16 mm and 500 mm long is subjected to a tensile force 40 kN. The modulus of
elasticity for steel may be taken as 200 kN/mm
2
. Find stress, strain and elongation of the bar due to applied load
Given that:
Load P = 40 kN = 40 ×1000 N
E = 200 kN/mm
2
= 200 ×10
3
N/mm
2
L = 500 mm
Diameter of the rod d = 16 mm
To find :
•Stress
•Stain
•Elongation in the bar
A = 201.06 mm
2
Stress
p = 198.94 N/mm
2
Strain
e = 0.0009947
Elongation in the bar
∆= 0.497 mm
2/2/2023 74
Stress and strain
Question:2 A specimen of steel 20 mm diameter with a gauge length of 200 mm is tested to destruction. It has an extension of
0.25 mm under a load of 80 kN and the load at elastic limit is 102 kN. The maximum load is 130 kN. The total extension at
fracture is 56 mm and diameter at neck is 15 mm. Find
(i)Thestressatelasticlimit.
(ii)Young’smodulus.
(iii)Percentageelongation.
(iv)Percentagereductioninarea.
(v)Ultimatetensilestress.
Answer:
(i) The stress at elastic limit = 324.675 N/mm
2
(ii) Young’s modulus = 203718 N/mm
2
(iii) Percentage elongation = 28
(iv) Percentage reduction in area = 43.75
(v) Ultimate tensile stress = 413.80 N/mm
2
2/2/2023 75
Question:2 A specimen of steel 20 mm diameter with a gauge length of 200 mm is tested to destruction. It has an extension of
0.25 mm under a load of 80 kN and the load at elastic limit is 102 kN. The maximum load is 130 kN. The total extension at
fracture is 56 mm and diameter at neck is 15 mm. Find
(i)Thestressatelasticlimit.
(ii)Young’smodulus.
(iii)Percentageelongation.
(iv)Percentagereductioninarea.
(v)Ultimatetensilestress.
Answer:
(i) The stress at elastic limit = 324.675 N/mm
2
(ii) Young’s modulus = 203718 N/mm
2
(iii) Percentage elongation = 28
(iv) Percentage reduction in area = 43.75
(v) Ultimate tensile stress = 413.80 N/mm
2
2/2/2023 75
Stress and strain
Question3:The bar shown in Fig. is tested in universal testing machine. It is observed that at a load of 40 kN the total extension
of the bar is 0.280 mm. Determine the Young’s modulus of the material.
∆ = ∆
1
+ ∆
2
+ ∆
3
∆
=
??
5
?
5
?
+
??
6
?
6
?
+
??
7
?
7
?
E = 200990 N/mm
2
2/2/2023 76
Question3:The bar shown in Fig. is tested in universal testing machine. It is observed that at a load of 40 kN the total extension
of the bar is 0.280 mm. Determine the Young’s modulus of the material.
∆ = ∆
1
+ ∆
2
+ ∆
3
∆
=
??
5
?
5
?
+
??
6
?
6
?
+
??
7
?
7
?
E = 200990 N/mm
2
2/2/2023 76
Stress and strain
Question 4: The bar shown a steel bar of cross–sectional area 250 mm
2
held firmly by the end supports and loaded by an axial
force of 25 kN. Take E
s
= 200 GPa
1.Find reaction at L and M
2.Find stress and strain in LN and NM
2/2/2023 77
R
L
R
L
R
L
R
M
25 kN
(25 –R
L
)
Question 4: The bar shown a steel bar of cross–sectional area 250 mm
2
held firmly by the end supports and loaded by an axial
force of 25 kN. Take E
s
= 200 GPa
1.Find reaction at L and M
2.Find stress and strain in LN and NM
2/2/2023 77
R
L
R
L
R
L
R
M
25 kN
(25 –R
L
)
Stress and strain
Question 5: A tapering rod has diameter d
1
at one end and it tapers uniformly to a diameter d
2
at the other end in a length L as
shown in Fig. If modulus of elasticity of the material is E, find its change in length when subjected to an axial force P.
Rate of change of diameter, k =
?5 ??6
?
Consider an elemental length of bar dx at a distance x from
larger end. The diameter of the bar at this section is
then d = d
1
+ kx
Cross-sectional area A =
πd
2
8
=
π(d
1
− kx)2
8
Extension of the element =
Extension of the entire bar
since d
1
–kL = d
2
∆
2/2/2023 78
Question 5: A tapering rod has diameter d
1
at one end and it tapers uniformly to a diameter d
2
at the other end in a length L as
shown in Fig. If modulus of elasticity of the material is E, find its change in length when subjected to an axial force P.
Rate of change of diameter, k =
?5 ??6
?
Consider an elemental length of bar dx at a distance x from
larger end. The diameter of the bar at this section is
then d = d
1
+ kx
Cross-sectional area A =
πd
2
8
=
π(d
1
− kx)2
8
Extension of the element =
Extension of the entire bar
since d
1
–kL = d
2
∆
2/2/2023 78
2/2/2023 79
Stress and strain
Question7:ThesteppedbarshowninFig.ismadeupoftwodifferentmaterials.Thematerial1hasYoung’smodulus=2×10
5
N/mm,whilethatofmaterial2is1×10
5
N/mm
2
.Findtheextensionofthebarunderapullof30kNifboththeportionsare
20mminthickness.
A
1
= 40 ×20 = 800 mm
2
A
2
= 30 ×20 = 600 mm
2
Stress and strain
Question7:ThesteppedbarshowninFig.ismadeupoftwodifferentmaterials.Thematerial1hasYoung’smodulus=2×10
5
N/mm,whilethatofmaterial2is1×10
5
N/mm
2
.Findtheextensionofthebarunderapullof30kNifboththeportionsare
20mminthickness.
A
1
= 40 ×20 = 800 mm
2
A
2
= 30 ×20 = 600 mm
2
2/2/2023 80
Stress and strain
Question8:Asteelflatofthickness10mmtapersuniformlyfrom60mmatoneendto40mmatotherendinalengthof600
mm.Ifthebarissubjectedtoaloadof80kN,finditsextension.TakeE=2×105MPa.Whatisthepercentageerrorifaverage
areaisusedforcalculatingextension?
Solution: If averages cross-section is considered instead of tapering
cross-section, extension is given by
Stress and strain
Question8:Asteelflatofthickness10mmtapersuniformlyfrom60mmatoneendto40mmatotherendinalengthof600
mm.Ifthebarissubjectedtoaloadof80kN,finditsextension.TakeE=2×105MPa.Whatisthepercentageerrorifaverage
areaisusedforcalculatingextension?
Solution: If averages cross-section is considered instead of tapering
cross-section, extension is given by
81
Stress and strain
Question9:Threepillars,twoofaluminiumandoneofsteelsupportarigidplatformof250kNasshowninFig.Ifareaofeach
aluminiumpillaris1200mm
2
andthatofsteelpillaris1000mm
2
,findthestressesdevelopedineachpillar.TakeE
s
=2×10
5
N/mm
2
andE
a
=1×10
6
N/mm
2
.
Solution:
Letforcesharedbyeachaluminium
pillarbeP
a
andthatsharedbysteelpillar
beP
s
.
The forces in vertical direction = 0
P
a
+ P
s
+ P
a
= 250
2P
a
+ P
s
= 250 (1)
From compatibility condition, we get
Δ
s
= Δ
aso, P
s
= 1.111 P
a
(2)
From equation (1) and (2), we get
Pa (2 + 1.111) = 250
∴ Pa = 80.36 kN
Hence from eqn. (1),
Ps = 250 –2 ×80.36
∴ Ps = 89.28 kN
∴Stresses developed are
σ
s
= 89.28 N/mm
2
σ
a
=
??????
??????
??????
??????
=
????????????.????????????×????????????????????????
????????????????????????
σ
a
= 66.97 N/mm
2
Stress and strain
Question9:Threepillars,twoofaluminiumandoneofsteelsupportarigidplatformof250kNasshowninFig.Ifareaofeach
aluminiumpillaris1200mm
2
andthatofsteelpillaris1000mm
2
,findthestressesdevelopedineachpillar.TakeE
s
=2×10
5
N/mm
2
andE
a
=1×10
6
N/mm
2
.
Solution:
Letforcesharedbyeachaluminium
pillarbeP
a
andthatsharedbysteelpillar
beP
s
.
The forces in vertical direction = 0
P
a
+ P
s
+ P
a
= 250
2P
a
+ P
s
= 250 (1)
From compatibility condition, we get
Δ
s
= Δ
aso, P
s
= 1.111 P
a
(2)
From equation (1) and (2), we get
Pa (2 + 1.111) = 250
∴ Pa = 80.36 kN
Hence from eqn. (1),
Ps = 250 –2 ×80.36
∴ Ps = 89.28 kN
∴Stresses developed are
σ
s
= 89.28 N/mm
2
σ
a
=
??????
??????
??????
??????
=
????????????.????????????×????????????????????????
????????????????????????
σ
a
= 66.97 N/mm
2
Stress and strain
Question10:Asteelboltof20mmdiameterpassescentrallythroughacoppertubeofinternaldiameter28mmandexternal
diameter40mm.Thelengthofwholeassemblyis600mm.Aftertightfittingoftheassembly,thenutisovertightenedby
quarterofaturn.Whatarethestressesintroducedintheboltandtube,ifpitchofnutis2mm?TakeE
s
=2×10
5
N/mm
2
andE
c
=1.2×10
5
N/mm
2
.
Solution:
Figureshowstheassembly.Lettheforce
sharedbyboltbeP
s
andthatbytubebeP
c
.
Sincethereisnoexternalforce,static
equilibriumconditiongives-
P
s
+ P
c
= 0 or P
s
= –P
c
(1)
Let the magnitude of force be P. Due to
quarter turn of the nut, the nut advances
by
5
8
×pitch =
5
8
×2 = 0.5 mm= ∆
Duringthisprocessboltisextendedand
coppertubeisshortenedduetoforceP
developed.LetΔ
s
beextensionofboltandΔ
c
shorteningofcoppertube.Finalpositionof
assemblybeΔ,then
Δ
s
+ Δ
c
= Δ(2)
-
Let, P
s
= P and P
c
= -P
-
P = 28.8168 KN
∴Stresses developed are
σ
s
=
?
?
?
?
=
6<.<5:×5444
?
?
σ
s
= 91.72 N/mm
2
σ
c
=
?
?
?
?
=
6<.<5:×5444
?
?
σ
s
= 44.96 N/mm
2
Question10:Asteelboltof20mmdiameterpassescentrallythroughacoppertubeofinternaldiameter28mmandexternal
diameter40mm.Thelengthofwholeassemblyis600mm.Aftertightfittingoftheassembly,thenutisovertightenedby
quarterofaturn.Whatarethestressesintroducedintheboltandtube,ifpitchofnutis2mm?TakeE
s
=2×10
5
N/mm
2
andE
c
=1.2×10
5
N/mm
2
.
Solution:
Figureshowstheassembly.Lettheforce
sharedbyboltbeP
s
andthatbytubebeP
c
.
Sincethereisnoexternalforce,static
equilibriumconditiongives-
P
s
+ P
c
= 0 or P
s
= –P
c
(1)
Let the magnitude of force be P. Due to
quarter turn of the nut, the nut advances
by
5
8
×pitch =
5
8
×2 = 0.5 mm= ∆
Duringthisprocessboltisextendedand
coppertubeisshortenedduetoforceP
developed.LetΔ
s
beextensionofboltandΔ
c
shorteningofcoppertube.Finalpositionof
assemblybeΔ,then
Δ
s
+ Δ
c
= Δ(2)
-
Let, P
s
= P and P
c
= -P
-
P = 28.8168 KN
∴Stresses developed are
σ
s
=
?
?
?
?
=
6<.<5:×5444
?
?
σ
s
= 91.72 N/mm
2
σ
c
=
?
?
?
?
=
6<.<5:×5444
?
?
σ
s
= 44.96 N/mm
2
2/2/2023 83
Stress and strain
Question 11: A bar of 25 mm diameter is tested in tension. It is observed that when a load of 60 kN is applied, the extension
measured over a guage length of 200 mm is 0.12 mm and contraction in diameter is 0.0045 mm. Find Poisson’s ratio and elastic
constants E, G, K.
Solution:
Now, P = 60 kN = 60000 N
Guage length L = 200 mm
E = 203718.3 N/mm
2
Stress and strain
Question 11: A bar of 25 mm diameter is tested in tension. It is observed that when a load of 60 kN is applied, the extension
measured over a guage length of 200 mm is 0.12 mm and contraction in diameter is 0.0045 mm. Find Poisson’s ratio and elastic
constants E, G, K.
Solution:
Now, P = 60 kN = 60000 N
Guage length L = 200 mm
E = 203718.3 N/mm
2
2/2/2023 84
Stress and strain
THERMAL STRESSES
Theconstantofproportionalityαiscalledcoefficientof
thermalexpansionandisdefinedaschangeinunitlengthof
materialduetounitchangeintemperature.
If the free expansion is prevented fully or partially the stresses
are induced in the bar, by the support forces.
If free expansion is permitted the bar would have expanded by
??????=??????????????????
Case 1:
Stress and strain
THERMAL STRESSES
Theconstantofproportionalityαiscalledcoefficientof
thermalexpansionandisdefinedaschangeinunitlengthof
materialduetounitchangeintemperature.
If the free expansion is prevented fully or partially the stresses
are induced in the bar, by the support forces.
If free expansion is permitted the bar would have expanded by
??????=??????????????????
Case 1:
2/2/2023 85
Stress and strain
Case 2: Expansion prevented
The expansion is prevented by developing compressive
force P at supports
Question 11. A steel rail is 12 m long and is laid at a temperature of
18°C. The maximum temperature expected is 40°C.
(i) Estimate the minimum gap between two rails to be left so that the
temperature stresses do not develop.
(ii) Calculate the temperature stresses developed in the rails, if:
(a) No expansion joint is provided.
(b) If a 1.5 mm gap is provided for expansion.
(iii) If the stress developed is 20 N/mm2, what is the gap provided
between the rails?
Where-E = 2 ×10
5
N/mm
2
and ??????= 12 ×10
–6
/°C.
Solution:
(i) The free expansion of the rails
= ?????? t L
= 12 ×10
-6
×(40 –18) ×12.0 ×1000
= 3.168 mm
Provide a minimum gap of 3.168 mm between the rails, so that
temperature stresses do not develop.
Stress and strain
Case 2: Expansion prevented
The expansion is prevented by developing compressive
force P at supports
Question 11. A steel rail is 12 m long and is laid at a temperature of
18°C. The maximum temperature expected is 40°C.
(i) Estimate the minimum gap between two rails to be left so that the
temperature stresses do not develop.
(ii) Calculate the temperature stresses developed in the rails, if:
(a) No expansion joint is provided.
(b) If a 1.5 mm gap is provided for expansion.
(iii) If the stress developed is 20 N/mm2, what is the gap provided
between the rails?
Where-E = 2 ×10
5
N/mm
2
and ??????= 12 ×10
–6
/°C.
Solution:
(i) The free expansion of the rails
= ?????? t L
= 12 ×10
-6
×(40 –18) ×12.0 ×1000
= 3.168 mm
Provide a minimum gap of 3.168 mm between the rails, so that
temperature stresses do not develop.
Stress and strain
(ii) (a) If no expansion joint is provided, free expansion
prevented is equal to 3.168 mm
Δ = 3.168 mm
(b) If a gap of 1.5 mm is provided, free expansion
prevented
Δ = α tL –δ = 3.168 –1.5 =1.668 mm
The compressive force developed is given by
??????=
(iii) If the stress developed is 20 N/mm
2
, then
(?????? )=
If δ is the gap, Δ = α t L –δ
δ = 3.168 –1.20 = 1.968 mm
(ii) (a) If no expansion joint is provided, free expansion
prevented is equal to 3.168 mm
Δ = 3.168 mm
(b) If a gap of 1.5 mm is provided, free expansion
prevented
Δ = α tL –δ = 3.168 –1.5 =1.668 mm
The compressive force developed is given by
??????=
(iii) If the stress developed is 20 N/mm
2
, then
(?????? )=
If δ is the gap, Δ = α t L –δ
δ = 3.168 –1.20 = 1.968 mm
Stress and strain
Question12:barofbrass20mmisenclosedinasteeltubeof40mmexternaldiameterand20mminternaldiameter.Thebar
andthetubesareinitially1.2mlongandarerigidlyfastenedatbothendsusing20mmdiameterpins.Ifthetemperatureis
raisedby60°C,findthestressesinducedinthebar,tubeandpins.
E
s
=2×10
5
N/mm
2
,
E
b
=1×10
5
N/mm
2
α
s
= 11.6 ×10
–6
/°C, α
b
= 18.7 ×10
–6
/°C.
Take:
Solution:
Given That:
t = 60°C
A
s
=
8
(40
2
–20
2
)
= 942.48 mm
2
A
b
=
8
(20
2
)
= 314.16mm
2
Horizontal equilibrium condition gives P
b
= P
s
, say P.
Δ
s
+ Δ
b
= α
b
t L –α
s
t L = (α
b
–α
s
)t L
where Δ
s
and Δ
b
are the changes in length of steels and brass bars
P = 11471.3 N
The pin resist the force P at the two cross-sections at junction of
two bars
Question12:barofbrass20mmisenclosedinasteeltubeof40mmexternaldiameterand20mminternaldiameter.Thebar
andthetubesareinitially1.2mlongandarerigidlyfastenedatbothendsusing20mmdiameterpins.Ifthetemperatureis
raisedby60°C,findthestressesinducedinthebar,tubeandpins.
E
s
=2×10
5
N/mm
2
,
E
b
=1×10
5
N/mm
2
α
s
= 11.6 ×10
–6
/°C, α
b
= 18.7 ×10
–6
/°C.
Take:
Solution:
Given That:
t = 60°C
A
s
=
8
(40
2
–20
2
)
= 942.48 mm
2
A
b
=
8
(20
2
)
= 314.16mm
2
Horizontal equilibrium condition gives P
b
= P
s
, say P.
Δ
s
+ Δ
b
= α
b
t L –α
s
t L = (α
b
–α
s
)t L
where Δ
s
and Δ
b
are the changes in length of steels and brass bars
P = 11471.3 N
The pin resist the force P at the two cross-sections at junction of
two bars
Stress and strain
Bending
M = bending moment at the section
I = moment of inertia about centroid axis
f = bending stress
y = distance of the fiber from neutral axis
E = modulus of elasticity and
R = radius of curvature of bent section.
ASSUMPTIONS
(i) The material is homogeneous and isotropic.
(ii) Modulus of elasticity is the same in tension and in
compression.
(iii) Stresses are within the elastic limit.
(iv) Plane section remains plane even after deformations.
(v) The beam is initially straight and every layer of it is
free to expand or contract.
(vi) The radius of curvature of bent beam is very large
compared to depth of the beam.
Bending
M = bending moment at the section
I = moment of inertia about centroid axis
f = bending stress
y = distance of the fiber from neutral axis
E = modulus of elasticity and
R = radius of curvature of bent section.
ASSUMPTIONS
(i) The material is homogeneous and isotropic.
(ii) Modulus of elasticity is the same in tension and in
compression.
(iii) Stresses are within the elastic limit.
(iv) Plane section remains plane even after deformations.
(v) The beam is initially straight and every layer of it is
free to expand or contract.
(vi) The radius of curvature of bent beam is very large
compared to depth of the beam.
Stress and strain
Question 1: A simply supported beam of span 3.0 m has a cross-section 120 mm ×180 mm. If the permissible stress in the
material of the beam is 10 N/mm
2
, determine.
(i) maximum udl it can carry
(ii) maximum concentrated load at a point 1 m from
support it can carry.
Neglect moment due to self weight
Solution:
Given that:
b = 120 mm, d = 180 mm,
I = bd
3
/12, and y = d/2
Section modulus Z = I/y
Z = bd
2
/6
Moment carrying capacity of the section
= fper ×Z
= 10 ×648000 N-mm
(i) maximum udl it can carry
Let maximum udl beam can carry be w/meter length as shown in
Fig.
In this case, we know that maximum moment occurs at mid span
and is equal to
w = 5.76 kN/m
Question 1: A simply supported beam of span 3.0 m has a cross-section 120 mm ×180 mm. If the permissible stress in the
material of the beam is 10 N/mm
2
, determine.
(i) maximum udl it can carry
(ii) maximum concentrated load at a point 1 m from
support it can carry.
Neglect moment due to self weight
Solution:
Given that:
b = 120 mm, d = 180 mm,
I = bd
3
/12, and y = d/2
Section modulus Z = I/y
Z = bd
2
/6
Moment carrying capacity of the section
= fper ×Z
= 10 ×648000 N-mm
(i) maximum udl it can carry
Let maximum udl beam can carry be w/meter length as shown in
Fig.
In this case, we know that maximum moment occurs at mid span
and is equal to
w = 5.76 kN/m
Stress and strain
(ii) maximum concentrated load at a point 1 m from support it can carry
Concentrated load at distance 1 m from the support
be P kN. Referring to Fig.
Equating it to moment carrying capacity, we get
P = 9.72 kN
(ii) maximum concentrated load at a point 1 m from support it can carry
Concentrated load at distance 1 m from the support
be P kN. Referring to Fig.
Equating it to moment carrying capacity, we get
P = 9.72 kN
Stress and strain
Question 2: A circular steel pipe of external diameter 60 mm and thickness 8 mm is used as a simply supported beam over
an effective span of 2 m. If permissible stress in steel is 150 N/mm
2
, determine the maximum concentrated load that can be
carried by it at mid span.
Solution:
External diameter D = 60 mm
Thickness = 8 mm
Internal diameter = 60 –2 ×8 = 44 mm
y
max
= 30 mm
Let maximum load it can carry be P kN
Equating maximum bending moment to moment carrying
capacity, we get
0.5P ×10
6
= 150 ×15073
P = 4.52 kN
Question 2: A circular steel pipe of external diameter 60 mm and thickness 8 mm is used as a simply supported beam over
an effective span of 2 m. If permissible stress in steel is 150 N/mm
2
, determine the maximum concentrated load that can be
carried by it at mid span.
Solution:
External diameter D = 60 mm
Thickness = 8 mm
Internal diameter = 60 –2 ×8 = 44 mm
y
max
= 30 mm
Let maximum load it can carry be P kN
Equating maximum bending moment to moment carrying
capacity, we get
0.5P ×10
6
= 150 ×15073
P = 4.52 kN
Stress and strainTorsion
Torsion Equation
Where J = Polar moment of inertia
τ = Shear stress induced due to torsion T.
G = Modulus of rigidity
ɵ= Angular deflection of shaft
R, L = Shaft radius & length respectively
Assumptions
The bar is acted upon by a pure torque.
1.Thesectionunderconsiderationisremotefromthe
pointofapplicationoftheloadandfroma
changeindiameter.
2.Adjacentcrosssectionsoriginallyplaneandparallel
remainplaneandparallelaftertwisting,
andanyradiallineremainsstraight.
3.ThematerialobeysHooke’slaw
4.Cross-sectionsrotateasifrigid,i.e.everydiameter
rotatesthroughthesameangle
Torsion Equation
Where J = Polar moment of inertia
τ = Shear stress induced due to torsion T.
G = Modulus of rigidity
ɵ= Angular deflection of shaft
R, L = Shaft radius & length respectively
Assumptions
The bar is acted upon by a pure torque.
1.Thesectionunderconsiderationisremotefromthe
pointofapplicationoftheloadandfroma
changeindiameter.
2.Adjacentcrosssectionsoriginallyplaneandparallel
remainplaneandparallelaftertwisting,
andanyradiallineremainsstraight.
3.ThematerialobeysHooke’slaw
4.Cross-sectionsrotateasifrigid,i.e.everydiameter
rotatesthroughthesameangle
Stress and strain
Maximum Principal Stress ( ??????
??????????????????
) & Maximum shear stress (
??????
??????????????????
)
Equivalent bending moment (M
e
) & Equivalent torsion (T
e
)
Maximum Principal Stress ( ??????
??????????????????
) & Maximum shear stress (
??????
??????????????????
)
Equivalent bending moment (M
e
) & Equivalent torsion (T
e
)
Stress and strain
Question 3: Determine the torsional stiffness of a hollow shaft of length L and having outside diameter equal to 1.5 times
inside diameter d. The shear modulus of the material is G.
Solution :
Question 3: Determine the torsional stiffness of a hollow shaft of length L and having outside diameter equal to 1.5 times
inside diameter d. The shear modulus of the material is G.
Solution :
Stress and strain
Solution :
Question4:Themaximumnormalstressandthemaximumshearstressanalyzedforashaftof150mmdiameterunder
combinedbendingandtorsion,werefoundtobe120MN/m
2
and80MN/m
2
respectively.Findthebendingmomentand
torquetowhichtheshaftissubjected.Ifthemaximumshearstressbelimitedto100MN/m
2
,findbyhowmuchthetorque
canbeincreasedifthebendingmomentiskeptconstant
Given that:
(i) Find the bending moment and torque to which the shaft is
subjected
Substituting the given values in the above equations, we have
Substituting this values in equation
M = 0.0265MNm
Substituting for M in equation
T = 0.0459MNm
Solution :
Question4:Themaximumnormalstressandthemaximumshearstressanalyzedforashaftof150mmdiameterunder
combinedbendingandtorsion,werefoundtobe120MN/m
2
and80MN/m
2
respectively.Findthebendingmomentand
torquetowhichtheshaftissubjected.Ifthemaximumshearstressbelimitedto100MN/m
2
,findbyhowmuchthetorque
canbeincreasedifthebendingmomentiskeptconstant
Given that:
(i) Find the bending moment and torque to which the shaft is
subjected
Substituting the given values in the above equations, we have
Substituting this values in equation
M = 0.0265MNm
Substituting for M in equation
T = 0.0459MNm
Stress and strain
(ii)Ifthemaximumshearstressbelimitedto100MN/m
2
,findbyhowmuchthetorquecanbeincreasedifthebending
momentiskeptconstant
Bending moment M to be kept cons tant = 0.0265MNm
T = 0.0607 MN-m
The increased torque = 0.0607 -0.0459
= 0.0148MNm
(ii)Ifthemaximumshearstressbelimitedto100MN/m
2
,findbyhowmuchthetorquecanbeincreasedifthebending
momentiskeptconstant
Bending moment M to be kept cons tant = 0.0265MNm
T = 0.0607 MN-m
The increased torque = 0.0607 -0.0459
= 0.0148MNm
Thermodynamics
2/2/2023 97
2/2/2023 97
First low of thermodynamics
2/2/2023 98
S.F.E.E. per unit mass basis
[h, W, Q should be in J/kg and C in m/s and g in m/s
2
]
[h, W, Q should be in kJ/kg and C in m/s and g in m/s
2
]S.F.E.E. per unit time basis
Where, w = mass flow rate (kg/s)
2/2/2023 98
S.F.E.E. per unit mass basis
[h, W, Q should be in J/kg and C in m/s and g in m/s
2
]
[h, W, Q should be in kJ/kg and C in m/s and g in m/s
2
]S.F.E.E. per unit time basis
Where, w = mass flow rate (kg/s)
First low of thermodynamics
2/2/2023 99
Question 1: A blower handles 1 kg/s of air at 20°C and consumes a power of 15 kW. The inlet and outlet velocities of air are
100 m/s and 150 m/s respectively. Find the exit air temperature, assuming adiabatic conditions. Take c
p
of air is 1.005 kJ/kg-K.
Solution:
From S.F.E.E. Answer: exit air temperature = 28.7
o
C
2/2/2023 99
Question 1: A blower handles 1 kg/s of air at 20°C and consumes a power of 15 kW. The inlet and outlet velocities of air are
100 m/s and 150 m/s respectively. Find the exit air temperature, assuming adiabatic conditions. Take c
p
of air is 1.005 kJ/kg-K.
Solution:
From S.F.E.E. Answer: exit air temperature = 28.7
o
C
First low of thermodynamics
2/2/2023 100
Question2:Aturbineoperatesundersteadyflowconditions,receivingsteamatthefollowingstate:Pressure1.2MPa,
temperature188°C,enthalpy2785kJ/kg,velocity33.3m/sandelevation3m.Thesteamleavestheturbineatthefollowing
state:Pressure20kPa,enthalpy2512kJ/kg,velocity100m/s,andelevation0m.Heatislosttothesurroundingsattherate
of0.29kJ/s.Iftherateofsteamflowthroughtheturbineis0.42kg/s,whatisthepoweroutputoftheturbineinkW?
Solution:
(Answer. power output of the turbine =112.51 kW)
2/2/2023 100
Question2:Aturbineoperatesundersteadyflowconditions,receivingsteamatthefollowingstate:Pressure1.2MPa,
temperature188°C,enthalpy2785kJ/kg,velocity33.3m/sandelevation3m.Thesteamleavestheturbineatthefollowing
state:Pressure20kPa,enthalpy2512kJ/kg,velocity100m/s,andelevation0m.Heatislosttothesurroundingsattherate
of0.29kJ/s.Iftherateofsteamflowthroughtheturbineis0.42kg/s,whatisthepoweroutputoftheturbineinkW?
Solution:
(Answer. power output of the turbine =112.51 kW)
First low of thermodynamics
2/2/2023 101
Question3:Anozzleisadeviceforincreasingthevelocityofasteadilyflowingstream.Attheinlettoacertainnozzle,the
enthalpyofthefluidpassingis3000kJ/kgandthevelocityis60m/s.Atthedischargeend,theenthalpyis2762kJ/kg.The
nozzleishorizontalandthereisnegligibleheatlossfromit.
(a)Findthevelocityatexistsfromthenozzle.
(b)Iftheinletareais0.1m2andthespecificvolumeatinletis0.187m
3
/kg,findthemassflowrate.
(c)Ifthespecificvolumeatthenozzleexitis0.498m
3
/kg,findtheexitareaofthenozzle.
Solution:
a.
Find V
2
i.e. Velocity at exit from S.F.E.E.
2/2/2023 101
Question3:Anozzleisadeviceforincreasingthevelocityofasteadilyflowingstream.Attheinlettoacertainnozzle,the
enthalpyofthefluidpassingis3000kJ/kgandthevelocityis60m/s.Atthedischargeend,theenthalpyis2762kJ/kg.The
nozzleishorizontalandthereisnegligibleheatlossfromit.
(a)Findthevelocityatexistsfromthenozzle.
(b)Iftheinletareais0.1m2andthespecificvolumeatinletis0.187m
3
/kg,findthemassflowrate.
(c)Ifthespecificvolumeatthenozzleexitis0.498m
3
/kg,findtheexitareaofthenozzle.
Solution:
a.
Find V
2
i.e. Velocity at exit from S.F.E.E.
First low of thermodynamics
2/2/2023 102
2/2/2023 102
First low of thermodynamics
2/2/2023 103
Question4:Aturbocompressordelivers2.33m3/sat0.276MPa,43°Cwhichisheatedatthispressureto430°Candfinally
expandedinaturbinewhichdelivers1860kW.Duringtheexpansion,thereisaheattransferof0.09MJ/stothe
surroundings.Calculatetheturbineexhausttemperatureifchangesinkineticandpotentialenergyarenegligible.
Solution:
= 1860 –(-90) kJ
= 1950 kJ
Answer. 157°C
2/2/2023 103
Question4:Aturbocompressordelivers2.33m3/sat0.276MPa,43°Cwhichisheatedatthispressureto430°Candfinally
expandedinaturbinewhichdelivers1860kW.Duringtheexpansion,thereisaheattransferof0.09MJ/stothe
surroundings.Calculatetheturbineexhausttemperatureifchangesinkineticandpotentialenergyarenegligible.
Solution:
= 1860 –(-90) kJ
= 1950 kJ
Answer. 157°C
First low of thermodynamics
2/2/2023 104
Question5:Aroomforfourpersonshastwofans,eachconsuming0.18kWpower,andthree100Wlamps.Ventilationairat
therateof80kg/henterswithanenthalpyof84kJ/kgandleaveswithanenthalpyof59kJ/kg.Ifeachpersonputsoutheat
attherateof630kJ/hdeterminetherateatwhichheatistoberemovedbyaroomcooler,sothatasteadystateis
maintainedintheroom.
Solution:
For steady state
= 1.9156 kW
Answer. 1.9156 kW
2/2/2023 104
Question5:Aroomforfourpersonshastwofans,eachconsuming0.18kWpower,andthree100Wlamps.Ventilationairat
therateof80kg/henterswithanenthalpyof84kJ/kgandleaveswithanenthalpyof59kJ/kg.Ifeachpersonputsoutheat
attherateof630kJ/hdeterminetherateatwhichheatistoberemovedbyaroomcooler,sothatasteadystateis
maintainedintheroom.
Solution:
For steady state
= 1.9156 kW
Answer. 1.9156 kW
First low of thermodynamics
2/2/2023 105
Question6:Thesteamsupplytoanenginecomprisestwostreamswhichmixbeforeenteringtheengine.Onestreamis
suppliedattherateof0.01kg/swithanenthalpyof2952kJ/kgandavelocityof20m/s.Theotherstreamissuppliedatthe
rateof0.1kg/swithanenthalpyof2569kJ/kgandavelocityof120m/s.Attheexitfromtheenginethefluidleavesastwo
streams,oneofwaterattherateof0.001kg/swithanenthalpyof420kJ/kgandtheotherofsteam;thefluidvelocitiesat
theexitarenegligible.Theenginedevelopsashaftpowerof25kW.Theheattransferisnegligible.Evaluatetheenthalpyof
thesecondexitstream.
Solution:
By mass balance
W
11
+ W
12
= W
21
+ W
22
∴W
22
= 0.01 + 0.1 –0.001 kg/s = 0.109 kg/s
29.522 + 257.62 = 0.42 + 0.109 ×h
22
+ 25
or 286.722 = 0.109 ×h
22
+ 25
or h
22
= 2401.2 kJ/kg
Answer. 1.9156 kW
2/2/2023 105
Question6:Thesteamsupplytoanenginecomprisestwostreamswhichmixbeforeenteringtheengine.Onestreamis
suppliedattherateof0.01kg/swithanenthalpyof2952kJ/kgandavelocityof20m/s.Theotherstreamissuppliedatthe
rateof0.1kg/swithanenthalpyof2569kJ/kgandavelocityof120m/s.Attheexitfromtheenginethefluidleavesastwo
streams,oneofwaterattherateof0.001kg/swithanenthalpyof420kJ/kgandtheotherofsteam;thefluidvelocitiesat
theexitarenegligible.Theenginedevelopsashaftpowerof25kW.Theheattransferisnegligible.Evaluatetheenthalpyof
thesecondexitstream.
Solution:
By mass balance
W
11
+ W
12
= W
21
+ W
22
∴W
22
= 0.01 + 0.1 –0.001 kg/s = 0.109 kg/s
29.522 + 257.62 = 0.42 + 0.109 ×h
22
+ 25
or 286.722 = 0.109 ×h
22
+ 25
or h
22
= 2401.2 kJ/kg
Answer. 1.9156 kW
First low of thermodynamics
2/2/2023 106
Question7:Thestreamofairandgasolinevapour,intheratioof14:1bymass,entersagasolineengineatatemperatureof
30°Candleavesascombustionproductsatatemperatureof790°C.Theenginehasaspecificfuelconsumptionof0.3
kg/kWh.Thenetheattransferratefromthefuel-airstreamtothejacketcoolingwaterandtothesurroundingsis35kW.
Theshaftpowerdeliveredbytheengineis26kW.Computetheincreaseinthespecificenthalpyofthefuelairstream,
assumingthechangesinkineticenergyandinelevationtobenegligible.
Solution:
In 1 hr. this m/c will produce 26 kWh for that we
need fuel = 0.3 ×26 = 7.8 kg fuel/hr.
∴Mass flow rate of fuel vapor and air mixture
2/2/2023 106
Question7:Thestreamofairandgasolinevapour,intheratioof14:1bymass,entersagasolineengineatatemperatureof
30°Candleavesascombustionproductsatatemperatureof790°C.Theenginehasaspecificfuelconsumptionof0.3
kg/kWh.Thenetheattransferratefromthefuel-airstreamtothejacketcoolingwaterandtothesurroundingsis35kW.
Theshaftpowerdeliveredbytheengineis26kW.Computetheincreaseinthespecificenthalpyofthefuelairstream,
assumingthechangesinkineticenergyandinelevationtobenegligible.
Solution:
In 1 hr. this m/c will produce 26 kWh for that we
need fuel = 0.3 ×26 = 7.8 kg fuel/hr.
∴Mass flow rate of fuel vapor and air mixture
Second low of thermodynamics
2/2/2023 107
Comparison of heat engine, heat pump and refrigerating machine
2/2/2023 107
Comparison of heat engine, heat pump and refrigerating machine
Second low of thermodynamics
2/2/2023 108
Question1:Aninventorclaimstohavedevelopedanenginethattakesin105MJatatemperatureof400K,rejects42MJat
atemperatureof200K,anddelivers15kWhofmechanicalwork.Wouldyouadviseinvestingmoneytoputthisenginein
themarket?
Solution:
Maximum thermal efficiency of his engine possible
That engine and deliver output = η ×input
= 0.5 ×105 MJ
= 52.5 MJ = 14.58 kWh
As he claims that his engine can deliver more work than ideally possible so I would not advise to investing money.
2/2/2023 108
Question1:Aninventorclaimstohavedevelopedanenginethattakesin105MJatatemperatureof400K,rejects42MJat
atemperatureof200K,anddelivers15kWhofmechanicalwork.Wouldyouadviseinvestingmoneytoputthisenginein
themarket?
Solution:
Maximum thermal efficiency of his engine possible
That engine and deliver output = η ×input
= 0.5 ×105 MJ
= 52.5 MJ = 14.58 kWh
As he claims that his engine can deliver more work than ideally possible so I would not advise to investing money.
Second low of thermodynamics
2/2/2023 109
Question 2: If a refrigerator is used for heating purposes in winter so that the atmosphere becomes the cold body and the
room to be heated becomes the hot body, how much heat would be available for heating for each kW input to the driving
motor? The COP of the refrigerator is 5, and the electromechanical efficiency of the motor is 90%. How does this compare
with resistance heating?
Solution:
(COP)
ref.
= (COP)
H.P
–1
∴(COP)
H.P.
= 6
But motor efficiency 90% so
= 0.1852 H
= 18.52% of Heat (direct heating)
2/2/2023 109
Question 2: If a refrigerator is used for heating purposes in winter so that the atmosphere becomes the cold body and the
room to be heated becomes the hot body, how much heat would be available for heating for each kW input to the driving
motor? The COP of the refrigerator is 5, and the electromechanical efficiency of the motor is 90%. How does this compare
with resistance heating?
Solution:
(COP)
ref.
= (COP)
H.P
–1
∴(COP)
H.P.
= 6
But motor efficiency 90% so
= 0.1852 H
= 18.52% of Heat (direct heating)
Second low of thermodynamics
2/2/2023 110
Question3:Ahouseholdrefrigeratorismaintainedatatemperatureof2°C.Everytimethedoorisopened,warmmaterial
isplacedinside,introducinganaverageof420kJ,butmakingonlyasmallchangeinthetemperatureoftherefrigerator.The
doorisopened20timesaday,andtherefrigeratoroperatesat15%oftheidealCOP.ThecostofworkisRs.2.50perkWh.
Whatisthemonthlybillforthisrefrigerator?Theatmosphereisat30°C.
Solution:
Ideal COP of Ref.
Actual COP = 0.15 ×COP ideal = 1.4732
Heat to be removed in a day (Q
2
) = 420 ×20 kJ
= 8400 kJ
∴Work required = 5701.873 kJ/day = 1.58385 kWh/day
Electric bill per month = 1.58385 ×0.32 ×30 Rupees
= Rs. 15.20
2/2/2023 110
Question3:Ahouseholdrefrigeratorismaintainedatatemperatureof2°C.Everytimethedoorisopened,warmmaterial
isplacedinside,introducinganaverageof420kJ,butmakingonlyasmallchangeinthetemperatureoftherefrigerator.The
doorisopened20timesaday,andtherefrigeratoroperatesat15%oftheidealCOP.ThecostofworkisRs.2.50perkWh.
Whatisthemonthlybillforthisrefrigerator?Theatmosphereisat30°C.
Solution:
Ideal COP of Ref.
Actual COP = 0.15 ×COP ideal = 1.4732
Heat to be removed in a day (Q
2
) = 420 ×20 kJ
= 8400 kJ
∴Work required = 5701.873 kJ/day = 1.58385 kWh/day
Electric bill per month = 1.58385 ×0.32 ×30 Rupees
= Rs. 15.20
Second low of thermodynamics
2/2/2023 111
Question4:AheatpumpworkingontheCarnotcycletakesinheatfromareservoirat5°Canddeliversheattoareservoirat
60°C.Theheatpumpisdrivenbyareversibleheatenginewhichtakesinheatfromareservoirat840°Candrejectsheattoa
reservoirat60°C.Thereversibleheatenginealsodrivesamachinethatabsorbs30kW.Iftheheatpumpextracts17kJ/s
fromthe5°Creservoir,determine
(a)Therateofheatsupplyfromthe840°Csource
(b)Therateofheatrejectiontothe60°Csink.
Solution:
COP of H.P.
Q
3
= W
H.P.
+ 17
Work output of the Heat engine
W
H.E.
= 30 + 3.36 = 33.36 kW
(b) Rate of heat rejection
to the 333 K
(i)From H.E. = Q
1
–W
= 47.61 –33.36 =
14.25kW
(ii) For H.P. = 17 + 3.36
= 20.36 kW
∴Total = 34.61 kW
2/2/2023 111
Question4:AheatpumpworkingontheCarnotcycletakesinheatfromareservoirat5°Canddeliversheattoareservoirat
60°C.Theheatpumpisdrivenbyareversibleheatenginewhichtakesinheatfromareservoirat840°Candrejectsheattoa
reservoirat60°C.Thereversibleheatenginealsodrivesamachinethatabsorbs30kW.Iftheheatpumpextracts17kJ/s
fromthe5°Creservoir,determine
(a)Therateofheatsupplyfromthe840°Csource
(b)Therateofheatrejectiontothe60°Csink.
Solution:
COP of H.P.
Q
3
= W
H.P.
+ 17
Work output of the Heat engine
W
H.E.
= 30 + 3.36 = 33.36 kW
(b) Rate of heat rejection
to the 333 K
(i)From H.E. = Q
1
–W
= 47.61 –33.36 =
14.25kW
(ii) For H.P. = 17 + 3.36
= 20.36 kW
∴Total = 34.61 kW
Second low of thermodynamics
2/2/2023 112
Question5:Aheatengineoperatingbetweentworeservoirsat1000Kand300Kisusedtodriveaheatpumpwhich
extractsheatfromthereservoirat300Kataratetwicethatatwhichtheenginerejectsheattoit.Iftheefficiencyofthe
engineis40%ofthemaximumpossibleandtheCOPoftheheatpumpis50%ofthemaximumpossible,whatisthe
temperatureofthereservoirtowhichtheheatpumprejectsheat?Whatistherateofheatrejectionfromtheheatpump
iftherateofheatsupplytotheengineis50kW?
Solution:
2/2/2023 112
Question5:Aheatengineoperatingbetweentworeservoirsat1000Kand300Kisusedtodriveaheatpumpwhich
extractsheatfromthereservoirat300Kataratetwicethatatwhichtheenginerejectsheattoit.Iftheefficiencyofthe
engineis40%ofthemaximumpossibleandtheCOPoftheheatpumpis50%ofthemaximumpossible,whatisthe
temperatureofthereservoirtowhichtheheatpumprejectsheat?Whatistherateofheatrejectionfromtheheatpump
iftherateofheatsupplytotheengineis50kW?
Solution:
Second low of thermodynamics
2/2/2023 113
Question 6: A reversible power cycle is used to drive a reversible heat pump cycle. The power cycle takes in Q
1
heat units at
T
1
and rejects Q
2
at T
2
. The heat pump abstracts Q
4
from the sink at T
4
and discharges Q
3
at T
3
. Develop an expression for the
ratio Q
4
/Q
1
in terms of the four temperatures.
Solution:
For H.E.
2/2/2023 113
Question 6: A reversible power cycle is used to drive a reversible heat pump cycle. The power cycle takes in Q
1
heat units at
T
1
and rejects Q
2
at T
2
. The heat pump abstracts Q
4
from the sink at T
4
and discharges Q
3
at T
3
. Develop an expression for the
ratio Q
4
/Q
1
in terms of the four temperatures.
Solution:
For H.E.
Second low of thermodynamics
2/2/2023 114
Question7:AreversibleengineoperatesbetweentemperaturesT1andT(T1>T).Theenergyrejectedfromthisengine
isreceivedbyasecondreversibleengineatthesametemperatureT.Thesecondenginerejectsenergyattemperature
T2(T2<T).
Show that:
(a) Temperature T is the arithmetic mean of temperatures T1 and T2 if the engines produce the same amount of work
output
(b) Temperature T is the geometric mean of temperatures T1 and T2 if the engines have the same cycle efficiencies.
Solution:
(b) If their efficiency is same then
(as T is + ve so –ve sign neglected)
∴T is Geometric mean of T
1
and T
2
.
2/2/2023 114
Question7:AreversibleengineoperatesbetweentemperaturesT1andT(T1>T).Theenergyrejectedfromthisengine
isreceivedbyasecondreversibleengineatthesametemperatureT.Thesecondenginerejectsenergyattemperature
T2(T2<T).
Show that:
(a) Temperature T is the arithmetic mean of temperatures T1 and T2 if the engines produce the same amount of work
output
(b) Temperature T is the geometric mean of temperatures T1 and T2 if the engines have the same cycle efficiencies.
Solution:
(b) If their efficiency is same then
(as T is + ve so –ve sign neglected)
∴T is Geometric mean of T
1
and T
2
.
Second low of thermodynamics
2/2/2023 115
Question 9: Two Carnot engines A and B are connected in series between two thermal reservoirs maintained at 1000 K and
100 K respectively. Engine A receives 1680 kJ of heat from the high-temperature reservoir and rejects heat to the Carnot
engine B. Engine B takes in heat rejected by engine A and rejects heat to the low-temperature reservoir. If engines A and B
have equal thermal efficiencies, determine
(a) The heat rejected by engine B
(b) The temperature at which heat is rejected by engine, A
(c) The work done during the process by engines, A and B respectively.
If engines A and B deliver equal work, determine
(d) The amount of heat taken in by engine B
(e) The efficiencies of engines A and B
Solution:
(Ans. (a) 168 kJ, (b) 316.2 K, (c) 1148.7, 363.3 kJ, (d) 924 kJ, (e) 45%, 81.8%)
2/2/2023 115
Question 9: Two Carnot engines A and B are connected in series between two thermal reservoirs maintained at 1000 K and
100 K respectively. Engine A receives 1680 kJ of heat from the high-temperature reservoir and rejects heat to the Carnot
engine B. Engine B takes in heat rejected by engine A and rejects heat to the low-temperature reservoir. If engines A and B
have equal thermal efficiencies, determine
(a) The heat rejected by engine B
(b) The temperature at which heat is rejected by engine, A
(c) The work done during the process by engines, A and B respectively.
If engines A and B deliver equal work, determine
(d) The amount of heat taken in by engine B
(e) The efficiencies of engines A and B
Solution:
(Ans. (a) 168 kJ, (b) 316.2 K, (c) 1148.7, 363.3 kJ, (d) 924 kJ, (e) 45%, 81.8%)
2/2/2023 116
Fluid Mechanics
Fluid Mechanics
Fluid Mechanics
2/2/2023 117
2/2/2023 117
2/2/2023 118
Flow between Plates
Flow between Plates
Second low of thermodynamics
2/2/2023 119
Question 9: Two Carnot engines A and B are connected in series between two thermal reservoirs maintained at 1000 K and
100 K respectively. Engine A receives 1680 kJ of heat from the high-temperature reservoir and rejects heat to the Carnot
engine B. Engine B takes in heat rejected by engine A and rejects heat to the low-temperature reservoir. If engines A and B
have equal thermal efficiencies, determine
(a) The heat rejected by engine B
(b) The temperature at which heat is rejected by engine, A
(c) The work done during the process by engines, A and B respectively.
If engines A and B deliver equal work, determine
(d) The amount of heat taken in by engine B
(e) The efficiencies of engines A and B
Solution:
(Ans. (a) 168 kJ, (b) 316.2 K, (c) 1148.7, 363.3 kJ, (d) 924 kJ, (e) 45%, 81.8%)
2/2/2023 119
Question 9: Two Carnot engines A and B are connected in series between two thermal reservoirs maintained at 1000 K and
100 K respectively. Engine A receives 1680 kJ of heat from the high-temperature reservoir and rejects heat to the Carnot
engine B. Engine B takes in heat rejected by engine A and rejects heat to the low-temperature reservoir. If engines A and B
have equal thermal efficiencies, determine
(a) The heat rejected by engine B
(b) The temperature at which heat is rejected by engine, A
(c) The work done during the process by engines, A and B respectively.
If engines A and B deliver equal work, determine
(d) The amount of heat taken in by engine B
(e) The efficiencies of engines A and B
Solution:
(Ans. (a) 168 kJ, (b) 316.2 K, (c) 1148.7, 363.3 kJ, (d) 924 kJ, (e) 45%, 81.8%)
2/2/2023 120
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