Unit:1 Energy principles
Part : B
6 marks
1.Find the deflection of a simply supported carrying a concentrated load at mid span. Assume uniform
flexural rigidly.
Mx= x
=
Deflection= dx
= dx
=
= [ ]
= [wx( )3]
=
2. Find the deflection at free end of the cantilever of length ’l’ carrying a u dl w/l over the whole span
P is a imaginary load at end B
Mx=-px- =-x
Deflection= dx
=
=
P=0
=
= [ ]
=
3. Define unit load method and explain it.
The principle of virtual work is based on the conservation of energy for a structure which implies loads
is equal to the internal energy stored in the structure
Ue=Ui
For beam, deflection =
In general,the principle of virtual work and energy states ΣP Δ = Σfδ
4.State Maxwell’s theorem and explain it.
It states that the deflection at any point P resting from the application of the load at any other point Q
is the same as the deflection of a resulting from the application of the same load at P.
5.Write down the steps to be followed in unit load method. 1.For the given load system,the forces in
the various members of the structures will be first determined.Let these forces be P1,P2 and P3.
2. The given load system is removed and a unit load is applied at the joint where the deflection is
desired.Now the forces be K1,K2 and K3 in the various members of the structures caused by the unit
load are computed.
3.The deflection at the joint is given by the summation Σ .This may be entered in a tabular form
as given below
Member P K L A
Deflection=
Part C (16 Marks)
1.A simply supported beam carrier a point load ‘P’ eccentrically on the span. Find the deflection under
the load assume uniform flexural rigidity
Taking moment about A
RBxl=Pa RA+RB=P
RB= PA=
=RAx=
=RBx =
=
=RBx=
= =
Total strain energy stored U= dx+ dx
= dx+ dx
= dx + dx
= [ ] + [ ]
= +
U= =
= =
2.Find the vertical and horizontal deflections of joint C of the pin jointed truss shown in figure.The areas
of horizontal member is 150 mm2 and the area of the members AC and BC are 200mm2 E=200 KN/mm2
=
=36ᵒ25’ AC= = 7.5m
Taking moment about A
12 RB -(9X6)=0
RB =4.5KN
RA+RB=9
RA=4.5KN
At joint A
ΣV=0
4.5+fAC =0
fAC=-7.5KN
ΣH=0
fAB + fAC =0
fAB =6.003KN
At joint B
ΣV=0
4.5+fBC =0
fBC =-7.5KN
T o find the vertical deflection C, remove the given load and apply 1KN vertical load
VA+VB=1KN
ΣH=0
VA X 12 – 1X6 =0
VA =0.5 KN VB=0.5 KN
Consider joint A
ΣV=0 ΣH=0
0.5+KAC =0 KAB + KAC =0
KAC =-0.833KN KAB=0.667
Consider joint B
ΣV=0
KBC =0
KBC =-0.833KN
Member P K L A
AC
-7.5 -0.833 7500 200 234.36
BC
-7.5 -0.833 7500 200 234.36
AB 6 0.666 7500 150 310.68
788.242
δ =
= X 788.242 = 3.9412 mm
To find the horizontal deflection at C apply a horizontal load of 1 KN at C
At joint A there will be a opposite force due to 1KN load at C
Moment about A
1X 4.5 + VB X 12 = 0
VA =0.37 KN VB=-0.37 KN
A t joint A
ΣV=0
0.375+KAC =0
KAC =-0.625
ΣV=0
-1+KAC KAB=0
KAB= 1.5KN
A t joint B
ΣV=0
-0.375+KBC =0
KBC =0.625
Member P K L A
AC
-7.5 -0.625 7500 200 0.878
BC
-7.5 -0.625 7500 200 -0.878
AB 0 1.5 12000 150 36
δ = 3.6mm
3.Calculate the central deflection of a simply supported beam loaded as shown in figure
Apply an imaginary load P at the centre of the beam
Taking moment about A
8RB=10 X 2 RA+RB= 10
RB=2.5KN RA = 7.5KN
Total reaction at end (A)[including imaginary]
RA =7.5 + , RB =2.5 +
BM at any section x distance from A (O-4m)
=RAx- 10(x-2)
=(7.5 + )x – 10x + 20
= + – 10x + 20
=
BM at any section x distance from B (O-4m)
= RBx
=(2.5 + )x
= +
=
Deflection= dx + dx
= dx + dx
= dx + dx
= [ ] + [ ]
P=0
= [ ] + [ ] = -
UNIT-2 Indeterminate beams
Part : B
6 Marks
1.The cantilever of length l carries a u dl of w/l is propped at the free end.If the prop holds the end at
the level of fixed end.Find the reactions of prop.
Let ax be the moment of area of cantilever beam about the prop
ax=
=
=
Let a’x’ be the moment of area of cantilever beam about the prop reaction only
a’x’= xl x pl x x l
a’x’=
Bm@A=
Now
a’x’=ax
=
P = wl BM@A=pl
2.A fixed beam of length 5m carries a u dl of 9 KN/m2.Find the fixing moments at the ends and deflection
at the centre
Given Data
L=5m
W=9KN/m
I=4.5 X 10-4 m4
E-1 X 107 KN/m2
Fixing moments at the ends
MA=MB= = = 18.75KNm
The deflection at the centre
Yc = =
=0.003254m
=-3.254mm
3.Find the fixing moments and support reaction of a fixed beam AB of 6m carrying a load of 4KN/m over
the left half of the span
Given Data
L=6m
W=4KN/m
The area of a’ BM diagram due to end moment is a’ =1/2(MA+MB) X 6
=3(MA+MB)
Taking moment about A
RB X 6 = 4 X 3 1.5 RB = 3KN
RA + RB = 12 RA = 9KN
MX=RA X x – 4 XxX
=9x – 2x2
Area=
=[ ]=22.5
A =Area + Area of triangle
=Area + X 9 X 3 = 22.5 + 0.5 X 9 X 3 = 36
MA+MB =12
ax = dx + Area of triangle
=
=
=[ ] +54
=94.5
a’x’ =MB X L X + ) L
=18 MB + 6 MA - 6 MB
=6(MA +2MB)
ax =a’x’
MA+2MB = =15.75
Solve equation 1 and 2,
MA=3.75KN
MB=8.25KN
4.A Continuous beam ABC covers two consecutive span AB and BC of lengths 4m and 6m Carrying a u dl
of 6kN/m respectively .If the ends A and C are simply supported .Find the support moments at A and C
are simply supported .
Given data
L1=4m
L2=6m
For span AB = = =12KN m
For span BC = = =45 KN m
End A and C are simply supported
MA = Mc =0
MAL1+2MB(L1 + L2)+MC
2L2 = +
4 + 2 MB(4+6)+ 0 L2 = +
= =96
= =540
20 MB =96+540
MB =31.8KN
Part-C(10 marks)
1. Calculate the fixing moment of fixed beam and deflection at the centre of the beam carrying a
concentrated point load at centre of the spam
a=a’
a= =
a’=-MA X l
MA= -
MB
B.M of fixed beam at any section at distance x from A
= - ’
= x
’=-
= - =
Deflection at
EI =
EI = x-
Integrate the above equation,
EI = - +
At x=0
EI = -
Integrate the above equation,
EIy= - +
At x=0, y=0
EIy= -
Differentiate at
EIy= -
y =
2.Find an expression for the deflection for a fixed beam carrying a u dl throughout the span
a=a’
a=
a’=- MA X l
= MA X l
MA=
Mx = - ’
= x - x
’ =
Mx =( x - x) -
EI = - ’
EI =( x - x) -
Integrate the above equation,
EI = - -
At x=0
EI = - -
Integrate the above equation,
EIy= - - +
At x=0, y=0
EIy= - -
Differentiate at
EIy= - –
y =
3.A cotinuous beam ABC of uniform section with span AB and BC as 4m is fixed at A and simply
supported at B and C .The beam is carrying a u dl of 4 KN/m run throughout its length.Find the support
moments and the reactions.Also draw the bending moment and shear force diagram
For span AB = = =12KNm
For span BC = = =12KNm
The extreme end C is simply supported hence span A’A and AB
M0 0+2MA(0+L1)+MBL1 =
2 MA (0+4)+4 =0+
= =64
8 MA+4 MB= X 64
2MA+ MB=24
For span AB and BC
MAL1+2MB(L1 + L2)+MCL2 = +
MA 4 + 2 MB(4+4)+ 0 L2 = +
= =64
= =64
4 MA+16 MB= X 64 + X 64
MA+4 MB=48
MA=6.86KNm
MB=10.28KNm
Find reactions
Taking moment about B
MB=RC X 4 – 6 X 4 X
RC=9.43KN
MB= MA+4 X RA– 6 X 4 X
RA= = 11.14 KN
RB=Total load – (RA+ RC)
=(6 X 4) - 11.14 – 9.43
=27.43
UNIT 3 Columns
Part : B (6 Marks)
1.A steel rod 5m long and of 40mm diameter is used as a column with one end fixed and other end is
free.Determine a crippling load by Euler’s formula.Take E=200Gpa
Given Data
L= 500mm
D=40mm
E=200Gpa=2 x 105 N/mm2
Cripping load
P= =
= 2480.42 N
I= =
=125.66 x 103 mm4
2.The external and internal diameter of a hollow cast iron column are 5 cm and 4cm respectively.If the
length of this column is 3m and both of its ends are fixed.Determine the crippling load.Using Rankine’s
formula.Take the values of =550N/mm2 and a= in Rankine’s formula.
Given Data
D=5 cm
d=4cm
A= (52-42)
= 7.0695cm2
A=706.9mm2
I = (54-44)
= 181132.66mm4
K=
Substitute the values of I and A, we get
K = 16.00mm
= = = 1500mm
P =
=
=
= 59881.822N
3.A 1.5m long column has a circular cross section of 5cm diameter one of the end of the column is fixed
and other end is free.Determine the crippling load Using Rankine’s formula.Take the values of
=560N/mm2 and a= in Rankine’s formula.
Given Data
d=5 cm
l=1.5m=1500mm
A= (52)
= 19.635cm2
A= 19.635 x 102mm2
I = (54)
= 30.7cm4
= 30.7x 104mm4
K=
Substitute the values of I and A, we get
K = 12.5mm
=2l =2 X 1500
=5000mm
P =
=
= 29708.1N
4.Define the term columns,strut and assumption made in Euler’s column theory for long columns
Member of the structure is vertical and both of its ends are fixed rigidly while subjected to axial
compressive load, the member is known as columns
Member of the structure is not vertical and one or both of its ends are hinged, the bar is known
as strut
Assumptions
1.The column is initially perfectly straight and the load is applied axially
2.The cross section of the column is uniform throughout its length
3.The column material is perffectly elastic,homogeneous and isotropic and obeys Hooke’s Law
4.The length of the column is very large as compared to its lateral dimensions
5.The direct stress is very small as compared to the bending stress
6.The column will fail by bucking alone
7.The self weight of column is negligible
PART-C
(10 Marks)
1.Find an expression for crippling load when both the ends of the column are hinged
M=-Py
EI =
EI = -Py
+ Y =0
The solution of the above differential equation is
Y=C1 ) +C2
(i)At A,x=0 and y=0
0= C1 + C2
C1 =0
(ii)At A,x=l and y=0
0=C1 ) +C2
0=0+ C2
C2 =0
=0
=0 or Π or 2Π
Taking least value
=Π
P=
2.Find the expression for crippling load when one end of the column is fixed and the other end is free
M=p(a-y)
EI = p(a-y)
=pa-py
EI + py=pa
+ y= a
The solution is
Y=C1 ) +C2
(i)At A,x=0 and y=0
0= C1 + C2 + a
= C1 X 1 + C2X 0 + a
C1 =-a
(ii)At A,x=0 and =0
Differentiating the equation w ith respect to x
= C1 +C2 )
0=- C1 X 0 +C2 X 1
C2 = 0
C2 =0
= 0
y=-a ) + a
At the free end x=l y=a
a=-a ) + a
0=-a )
) =0
) = or
= or
Taking least values
=
P=
3.Find the expression for crippling load when both the end of the column is fixed
M = -py
EI = -py
EI +py =
+ y= a
= X
= X
The solution of the above differential equation is
Y=C1 ) +C2 +
(i)At A,x=0 and y=0
0= C1 + C2 +
= C1 X 1 + C2X 0 +
C1 =-
Differentiating the equation of y with respect to x
= C1 +C2 ) +0
=- C1 +C2 )
(ii)At A,x=0 and =0
0=- C1 X 0 +C2 X 1
C2 = 0
C2 =0
= 0
y=- ) + a
(iii)At x=l, y=0
0=- ) +
) =
) =1
) = or or
= or or
Taking least values
=
P=
4. Find the expression for crippling load when one end of the column is fixed and the other end is hinged
M=-py+H(l-x)
EI = -py+H(l-x)
EI +py =H(l-x)
+ y =
=
The solution of the above differential equation is
Y=C1 ) +C2 +
(i)At A,x=0 and y=0
0= C1 + C2 +
= C1 X 1 + C2X 0 +
C1 =-
Differentiating the equation of y with respect to x
= C1 +C2 ) -
=- C1 +C2 ) -
(ii)At A,x=0 and =0
0=- C1 X 0 +C2 X 1 -
C2 = 0
C2 =
y= ) + +
(iii)At the end x=l, y=0
0=- ) + +
0=- ) +
= )
= )
=
=4.5radians
=4.52
P =
P=
5. Determine the maximum and minimum hoop stress across the section of a pipe of 400mm internal
diameter and 100mm thick, when the pipe contains a fluid at a pressure of 8N/mm2
Given Data
Internal diameter = 400mm
r1= =200mm
Thickness =100mm
External diameter=400 + 2 X 100 = 600mm
r2= = 300mm
P0=8 N/mm2
At x= r1,Px= P0=8 N/mm2
Px= - a
1.At x= r1=200mm ,Px=8 N/mm2
2.At x= r2= 300mm,Px=0
8= – a = – a
0 = – a = – a
Solve the above equations
8= -
8=
b= =576000
Substituting the values in equation
0= - a
a= 6.4
= + a = + 6.4
At x=200mm, = + 6.4 =20.8 N/mm2
At x=300mm, = + 6.4 =12.8 N/mm2
UNIT 4 State of stress in 3 D
Part : B (6 Marks)
1.A 5mm thick aluminium plate has a width of 300mm an da length of 600mm subjected to pull of
15000N and 9000N respectively in axial and transverse direction.Determine the normal,tangential and
resultant stress on a plane 50ᵒ
Given Data
b=300mm
l=600mm
t=5mm
P1=15000N
P2=9000N
Axial stress= = = 10 N/mm2
Transverse stress= = = 3 N/mm2
Normal stress = +
= + = 5.89 N/mm2
Tangential stress =
= = 3.447 N/mm2
Resultant stress =
=6.82 N/mm2
2.Explain the term maximum principal stress theory
According to Rankine’s theory, failure will occur if maximum principal stress
exceeds a value of ,the direct stress at a elastic limit,
For no failure maximum principal stress
In the case of two dimensional stress
and in the case of three dimensional stress
3.Explain the term maximum strain theory
According to this theory failure will occur when maximum principal strain occur
caused by the direct strain at a elastic limit
For no failure maximum principal strain
-
-poisson’s ratio
4.Explain the term maximum shear stress theory
According to this theory, failure will occur if maximum shear stress occur
exceeds the maximum shear caused by the direct stress at a elastic limit
For no failure maximum principal strain
In the case of two dimensional stress
in the case of three dimensional stress
5.Explain the term total strain energy theory
According to this theory, failure will occur if total strain energy occur exceeds the total strain
energy caused by the direct stress at a elastic limit
For no failure total strain energy
6.Explain the term shear strain energy theory
According to this theory, failure will occur if shear strain energy occur exceeds the shear strain
energy caused by the direct stress at a elastic limit
For no failure shear strain energy
Part- C
10Marks
1.The state of stress at a point is given by KN/m2
Determine the principal stress & direction associated with the normals to the surface of each
principle strain .
Let be the principal stresses ,for the existence of principal stress & principal planes.
I1 = =3000+0+0=3000
I2 = =0+0+0+ -
I2=
I3=
=0-3000
=
Cubic equation
-3000
mm2
=1000
Direction Cosines
=
A1= =
B1= =
C1= =
K1 =
= =0.8165
= =-0.4082
= =0.4082
Direction cosines
=1000 mm2
=
A1= =
B1= =
C1= =
K2 =
= =-0.577
= =-0.577
= =0.577
Direction cosines
=1000 mm2
=
A1= =
B1= =
C1= =
K3 =
2. A certain steel has proportionality limit of 280 mm2 in simple tension.Under a certain three
dimensional stress system,the principal stress are 129 mm2 tensile, 60N/ mm2tensile,&
30 mm2Compressive. Calculate the FOS according to Maximum principal stress theory
,Maximum shear stress, Total strain energy theory , Shear strain energy theory, Maximum strain
theory
Given Data
mm2
=60N/ mm2
Maximum principal stress theory
120
Factor of safety = =2.33
Maximum shear stress
=120 + 30 =150
Total strain energy theory
=133.487
FOS= =2.1
Shear strain energy theory
=130.766
FOS= =2.14
Maximum strain theory
120-(60-30) X0.3
=111
FOS= =2.52.
2.State and explain theories of failure
Maximum principal stress:
According to Rankine’s theory,failure will occur if maximum principal stress exceeds a value of ,the
direct stress at a elastic limit,
For no failure maximum principal stress
In the case of two dimensional stress
and in the case of three dimensional stress
Maximum strain theory
According to this theory failure will occur when maximum principal strain occur
caused by the direct strain at a elastic limit
For no failure maximum principal strain
-
-poisson’s ratio
Maximum shear stress theory
According to this theory, failure will occur if maximum shear stress occur
exceeds the maximum shear caused by the direct stress at a elastic limit
For no failure maximum principal strain
In the case of two dimensional stress
in the case of three dimensional stress
Total strain energy theory
According to this theory, failure will occur if total strain energy occur exceeds the total strain
energy caused by the direct stress at a elastic limit
For no failure total strain energy
Shear strain energy theory
According to this theory, failure will occur if shear strain energy occur exceeds the shear strain
energy caused by the direct stress at a elastic limit
For no failure shear strain energy
Unit 5 Advanced topics in bending of stress
Part : B (6 Marks)
1. A steel plate of width 120mm and of thickness 20mm is bent into a circular arc of radius 10m.
Determine the maximum stress induced and the bending moment which will produce the
maximum stress take E=2x10^5N/mm.
Given data:
b=120mm,t=20mm
Moment of inertia, I= bt3/12
= 120 x 20 3/12 = 8 x10 4 mm 4
Radius of curvature ,R =10m=10x10 3mm
E = 2x10 5 N/mm 2
=
σ = x y
y = t/2 = 20/2 = 10mm.
σ =200 N/mm2
M = = =16 N mm.
2. Calculate the maximum stress induced in a cast iron pipe of external dia 40mm of internal dia
20mm and length 4m when the pipe is supported at its ends and carries a point load of 80N at
its centre.
Given data:
D = 40mm
d = 20mm
L = 4000mm
W = 80 N
Maximum bending moment : = = Nmm.
I = -
= -
= -
= 117809.7mm 4
=
Y = = = 20mm
σ = x y
=
= 13.58 N/mm2
3. A rectangular beam 200mm deep & 300mm wide is simply supported over a span of 8m.what
udl/m. of the beam may occur ,if the bending stress is not to exceed 120N/mm^2.
Given data:
D = 200mm
B =300mm
L = 8m
σ = 120 N/mm2
Z = = = 2000000 mm 3
M = = = 8w Nm
M = σ Z
8000 w = 120 x2000000
W= 30 x1000
= 30 KN/m.
4. Determine the position of neutral axis on a curved beam of diameter 100mm to pure bending
moment of 11.5 KNm.The radius of curvature is 100mm.
Given data:
diameter 100mm
bending moment of 11.5 KNm.
The radius of curvature is 100mm.
= +
= +
=- =-
=-6.57mm
Part-C
10 marks
1. Determine : (1) location of neutral axis
(2)Maximum and minimum stress
(3) Ratio of maximum and minimum stress when curved beam of rectangular section
of width 20mm and of depth 40mm is subjected to pure bending of moment 600Nm.The beam is
curved in a plane parallel to depth.The mean radius of curvature is 50mm,Also calculate the
variation of the stresses across the section.
M=600Nm=600Nmm
b=20mm
d=40mm
R=50mm
=
=
=147.806
(1) location of neutral axis
=- =-
=-2.79mm
(2)Maximum and minimum stress
=
= =
=-154.140 N/mm2
Maximum stress will occur at extreme top layer
y=20mm
= = 87.488 N/mm2(tensile)
(3)Ratio of maximum and minimum stress
= =1.762
(4) variation of the stress
=
15[1+16.91( )]
N/mm2
5= 15[1+16.91( )]=38.059N/mm2
10=57.27 N/mm2
15=73.53N/mm2
20=87.48N/mm2
-5=-13.83N/mm2
-10=-48.41N/mm2
-15=-93.707N/mm2
-20=-154.140N/mm2
2.Determination of (1) Neutral axis
(2) Maximum and minimum stress when a curved beam of trapezoidal section of
bottom width 30mm,top width 20mm and height 40mm is subjected to pure bending of beam of B.M
600Nm.The bottom width is towards the centre of curvature .The radius of curvature is 50mm and the
beam is curved in a parallel to depth.Also calculate the variation of stresses.
=( )
=[ ]
=40-18.667=21.333mm
A= mm2
=
=
=140.59
(i)Location of neutral axis
=- =-
=-2.662mm
(2)Maximum and minimum stress
=
= =-115.126 N/mm2
=12[1+17.782( )]=75.802 N/mm2
Maximum stress will occur at extreme top layer
variation of the stress
=
= [1+17.782( )]
=12 N/mm2
=31.399 N/mm2
10=47.564 N/mm2
15=61.242N/mm2
21.33=75.815N/mm2
-5=-11.709N/mm2
-10=-41.436N/mm2
-15=-79.450N/mm2
-18.667=-115.126 N/mm2
3. A water main of 500mm internal diameter and 20mm thick is running full.The water main is of cast
iron and is supported at two points 10m apart.Find the maximum stress in the metal.The cast iron and
water weight 72000N/mm2and 10000M/mm2respectively.
=500mm=0.5m
t=20mm
= +2 X t =500 + 2 X 20 =540mm=0.54m
Internal area=
Area of pipe section= = =0.0327 m2
Moment of inertia of the pipe section
I= = =1.105 X mm4
Weight of the pipe for one meter= weight density of cast iron X volume of pipe
=72000x(Area of pipe section X length) (length=1m)
=72000x0.0327x1
=2376N
Weight of water for one metre run=weight density of water X volume of water
=10000XArea of water section X length
=10000X0.196X1
=1960N
Total=2376 X 1960=4336N
Maximum bending moment due to u dl
M= = =54200 Nmm
=
X y
y = = = 270mm
= X 270
= 13.18 N/mm2
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