SUBJECT-ANALYSIS OF STRUCTURE (CET-401)
Transcript of SUBJECT-ANALYSIS OF STRUCTURE (CET-401)
SUBJECT-ANALYSIS OF STRUCTURE
(CET-401)
BRANCH-CIVIL ENGG.
SEMESTER-4TH
ER.DEEPALI BARIK
4TH SEM CIVIL SUB – ANALYSIS OF STRUCTURE CET -401
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CHAPTER – 1
Short questions (2 marks)
Q1. Define statically determinate and statically indeterminate structure
2015, 2(a)
Ans: Statically determinate Structure : - Conditions of equilibrium are sufficient to fully
analyse the structure, i.e. called as statically determinate structure.
Statically Indeterminate structure:- Conditions of equilibrium are insufficient analyse
the structure fully, i.e. called as statically indeterminate structure.
Q.2. What is a deficient frame ? Write down the equation 2013(s), 1(a),
2016(s)
Ans: A deficient frame is an imperfect frame. In which the number of members are less
than (2J-3)
Equation n < (2J – 3)
Where, n = number of members
J = number of joints.
Q.3. What do you mean by a portal frame 2016, 7(a)
Ans: The portal frame is an example for a statically indeterminate structure. This frame can
be analysed by moment distribution method, slope deflection method etc.
Q.4. How a truss differs from a beam? 2015, 1(a) , 2013, 1(b)
Ans: The only difference between truss and beam is that a
Truss transmits the force only in the axial direction.
Beam transmits force both in axial & vertical direction.
Q.5. Explain how method of Joint differs over the method of section in plane
truss ? 2014, 7(a)
Ans: Only two unknown forces can be found by method of joints where as three unknown
forces can be found by method of sections.
Q.6. Write down the equation for the maximum deflection of a simple
supported beam of Span ‘l’ which a central point load (w) . 2017 (s)
2(a)
Ans:- Maxm Deflection , Yc= 𝑤𝑙3
48𝐸𝐼
Where W=point load
4TH SEM CIVIL SUB – ANALYSIS OF STRUCTURE CET -401
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L= span of beam
E= Modulus of elasticity
I = moment of inertia
Q.7.Write down the statement of second the orem of moment area
method. 2017 3(a)
Ans: Mohr’s theorem-II
It states ‘the intercept taken on a vertical reference line of tangents at any
two points on an elastic curve, is equal to the moment of the B.M. diagram
between these points about the reference line divided by EI.’
5 MARKS
Q.1. By using method of section, find out the axial forces along member1, 2 and
3 as shown in the figure. 2014, 7(c)
Ans: Considering RHS of the section X – X
ME = 0 FCB × EC + 6 × 4 = 0
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Q.2. Find the forces in various member of the truss as shown in figure and
tabulate the results 2015 1(b)
Ans:
Let Q be the inclination of the member BD with vertical.
= 30o, DA = 3 m
Joint A, resolving vertically
PAB sin 60o = 10000 N = 10 kN
PAB = 11.55 kN (Tensile)
Resolving horizontally,
PAD + PAB cos 60o = 0.
AAD = –5.78 kN (compressive)
Joint B,
CB
B EA
EA
EA EB
EB
EB
EB
( )6 4or F ( )6 / tan 30 3.46 kN
4 / tan30
(compress)
4M 0 F cos 30 6 4 6 8
cos 30
72or F 18 kN (Tension)
4
V 0 F sin 30 F sin30 12 0
or 18 0.5 F 0.5 12 0
or 9 12 E 0.5
3or F ( )6kN(compress)
0.5
4TH SEM CIVIL SUB – ANALYSIS OF STRUCTURE CET -401
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Resolving vertically,
PBA cos 30o + PBP cos 30o = 0.
PBA cos 30o = – PBD cos 30o.
PBA = – PBD {We know PBA = PAB)
PBD = –11.55 kN (compressive)
Resolving horizontally,
PBC = PBD sin 30o + PBA sin 3oo
PBC = –11.55 × ½ + 11.55 × ½ = 0.
Q.3. Find out forces in all the members with their nature as tensile or
compressive as shown in figure below. 2015 (s) 3(c)
Ans:
Member Tensile compressive
AB
AD
BC
BD
11.55
0
5.78
0
11.55
D
D
D
A
Support reactions,
R 6 4 4.5 2 1.5
R 6 21
21R 3.5kN
6
Member forces R (2 4) 3.5 2.5kN
Jo int D.
4TH SEM CIVIL SUB – ANALYSIS OF STRUCTURE CET -401
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DC D
DDC
DE DC
DE
V 0
F sin60 R 0
RF 4.04kN(compression)
sin60
H 0
F F cos 60 0
F 2kN(Tension)
Jo int A
A AB
AAB
o
AE AB
AE
V 0
R F sin60 0
RF 2.89kN(compression)
sin60
H 0
F F cos 60 0
F 1.45kN(Tension)
Jo int E
EB EC
EB EC
o o
EA EB CD EC
o
EA ED EC
BC
V 0
F sin60 F sin60 0
F F
H 0
F F cos 60 F F cos 60 0
F F F cos 60 0.
F 0.55(comprssion)
Jo intB
4TH SEM CIVIL SUB – ANALYSIS OF STRUCTURE CET -401
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7 MARKS
Q.1. The girder is loaded at ‘B’ & ‘C’ as shown in the fig. find the forces in all
members of the girder & indicate the nature of forces.2014, 1(c)
BC AB BE
BC
H 0
H F cos 60 F cos 60 0
F 1.17(Tensile)
D
D
A
o
CD D
CD
CD
o
CD ED
ED
ED
Support reactions
R 8 6 2 4 6 36
36or R 4.5 kN
8
R (6 4) 4.5 5.5 kN.
Member forces
Jo int D
V 0 F sin60 R 0
or F 0.866 4.5
4.5or F ( )5.2 kN (compression)
0.866
H F cos 60 F 0
or 5.2 0.5 F 0
or F 2.6 kN(
o
AB A
AB
AB
o
AB AE
AB
AE
Tension)
Jo int A
V 0 F sin60 R 0
or F 0.866 5.5 0
5.5or F ( )6.35 kN (comprssion)
0.866
H F cos 60 F 0
or 6.35 0.5 F 0
or F 3.18 kN (Tension)
B C
DAo60
o60 o60 o60
aR bRE8m
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Q.2. Find the force in the member BC shown in following fig. Find out the force on
members using method of section. 2016(s) 1(c)
Ans: Section 1 -1 as shown in fig.
To find the forces in BC. Consider the left part of section 1-1 & take moments about A
Q.3. Find out the force in members using method of section 2013, (3)
Ans:
o 0
EB EC
EB EC
EB EC
o 0
AB ED FB EC
0
EC
EC
o o
BC BE AB
BC
B
V 0 F sin60 F sin60 0
or F F 0
or F F 0
H 0 F F F cos 60 F cos 60 0
or 3.18 2.6 2F cos 60 0
0.58or F 0.58 kN (Tension)
2 0.5
Jo int B
H 0
F F cos 60 F cos 60 0
F ( 0.58) 0.5 6.35 0.5 0
F
C 2.89kN (compression)
B C
A
5m
o60o30
10kN
A B
C
5m
o60o30
10kN
A B
C
o60
10kN
PCB
o
CB
o
CB
CB
CB
V 0
10 AC cos 60 P AC 0
10AC cos 60P
AC
1P 10 5kN
2
P 5 kN (compression)
4TH SEM CIVIL SUB – ANALYSIS OF STRUCTURE CET -401
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Taking moments of the forces acting in the left part of the truss only about the
joint C and equating the same, PAB × l sin 60o = R1 × l
Now taking moments of the forces acting in the left part of the truss only about
the joint A and equating the same.
Taking moments of the force acting in the right part of the truss only about
the joint B and equating the same.
Now taking moments of the forces acting in the left part of the truss only
about the joint A and equating the same
1 1AB 1o o
R l R lP 1.16R KN(compression)
l sin60 lsin60
o
BC 1
11
BC 1oo
l lP tan60 R
4 4
lR
R4P 0.57 R KN(Tension)l tan60
tan604
o
AC 2
1 2AC 2o o
P lsin 30 R l
R l RP 2 R KN(compression)
lsin30 sin30
o
BC 2
22
BC 2oo
3l 3lP tan30 R
4 4
3lR
R4P 1.57 R KN(Tension)3l tan30
tan304
4TH SEM CIVIL SUB – ANALYSIS OF STRUCTURE CET -401
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CHAPTER:2
2 MARKS
Q.2. State the slope and deflection equation for a member by (i) moment area method (ii)
Double integration method. 2014, (2a)
Ans:i) Moment area method
ii) Double integration method
Q.3. Write down the relationship between the slope deflection and radius of
curvature. 2016, 3(a) , 2013 , 1(e)
Ans: The relation between the slope, deflection and radius of curvature is
Q.4. Write down the equation of deflection at free end of a cantilever beam of
span ‘l’ subjected to a point load ‘w’ at free end.2016, 4(a) 2013 1(c)
Ans: Equation of deflection,
Where w = point load
L = span of beam
E = Modulus of elasticity
I = M.I. of the beam
Q.5. Write down the statement of second theorem of moment area method.
2016, 5(a), 2013 1(g)
Ans: Moment area method
x2
x
x1
x
x x
x1
MdSlope equation
EI
M dDeflectionequationz
EI
x
dySlopeequation EI. Md f '(x)
dx
DeflectionequationEI.Y f '(x)dx f(x)
2
2
d yM EI
dx
3
B
WlY
3EI w
B
Byl
A
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Mohr’s theorem-II It states the intercept taken on a vertical reference line of
tangents at any two points on an elastic curve, is equal to the moment of the B.M.
diagram between these points about the reference line divided by EI.
Q Write down the equation for the maximum deflection of a simple
supported beam of Span ‘l’ which a central point load (w) . 2017 (s)
2(a)
Ans:- Max3 Deflection , Yc= 𝑤𝑙3
48𝐸𝐼
Where W=point load
L= span of beam
E= Modulus of elasticity
I = moment of inertia
Q. Write down the statement of second the orem of moment area
method. 2017 3(a)
Ans: Mohr’s theorem-II
It states ‘the intercept taken on a vertical reference line of tangents at any two
points on an elastic curve, is equal to the moment of the B.M. diagram
between these points about the reference line divided by EI.
Q) Calculate the maxn slope & deflection in case of a S/s beam of span
6m subjected to a point load 10kn at the middle of the span EI
constant?2017 5 (b)
ANS.
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In this case as the load is symmetrically placed the deflection will be maxm at
the mid span & slope shall be maxm at the ends.
Qmax = 𝐴
𝐸𝐼
Area of shaded.
∆=1
2 ×
𝑤𝑙
4×
𝑙
2
=1
2×
10×6
4×
6
2
=360
16 =22.s
Qmax = 22.5
EI
Ymax = 𝐴𝑥
𝐸𝐼
2
.16 3
Wl l
EL
=
3
3
48
10 (6) 45
48
Wl
EI
x
xEI EI
4TH SEM CIVIL SUB – ANALYSIS OF STRUCTURE CET -401
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5 MARKS
Q.1. A cantilever beam of 3 m long carried a point load of 40 kN at its free end.
Find the slope and deflection of the cantilever under the load. Take flexural
rigidity for the cantilever beam as 25 × 1012 N-mm2 2014, 1(b)
Ans: Given data
Length, L = 3m = 3000 mm
Point load, W = 40 kN = 40,000 N.
Flexural rigidity EI = 25 × 1012 N – mm2
Q.2. A simply supported beam of span 8 meters is subjected to a point load of 60
kN at the centre. Determine the maximum deflection at the centre by using
moment area method. Take EI of the beam section as 10 × 102 N-mm2.
2014, 2(b)
Ans: Given data
Span (l) 8 m = 8 × 103 mm
Point load (w) = 60 kN = 60 × 103 N
Flexural rigidity (EI) = 10 × 1012 N-mm2
Maximum deflection of the beam at its centre
2
B
2 11
12 13
3 3
B 12
WL(i) Slopeat the free end is i
2EI
40,000 3000 3.6 10
2 25 10 5 10
0.0072 rad.
(ii) DEflection at the free end is
WL 40,000 3000y 14.4mm
3EI 3 25 10
3
3 33
c 12
60 10 8 10WLy 64mm
48EI 48 10 10
4TH SEM CIVIL SUB – ANALYSIS OF STRUCTURE CET -401
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Q. A cantilever beam 120 mm wide and 160 mm deep is 2 m long. Determine the slope and deflection at the free end of the beam. When it carries a point
load of 30 kN at its free end. Take E for the cantilever beam as 200 GPa. 2014, 6(b)
Ans: Width, b = 120 mm
Depth, d = 160 mm Span, 1 = 2 m = 2000 mm Point load , w = 30 kN = 30 × 103 N
E = 200 GPa = 200 × 103 N/m3 Slope at the free end We know that moment of inertia of the beam section
Q.4. Derive the slope and deflection of simply supported beam with a central
point load ‘W’ with span length ‘L’ by double integration method 2014,7(b)
Ans: Simply supported beam with a central load
3 36 4
2 3 2
B 3 6
3 3 3
B 3 6
bd 120 160I 40.96 10 mm
12 12
and slope at the free end
wl 30 10 2000i 0.0073rad
2EI 2 200 10 40.96 10
Deflection of the free end
wl 30 10 2000y 9.76mm
3EI 3 200 10 40.96 10
x
2
2
2
1
WM x
2
d y WEI x
dx 2
dy WxEI C
dx 4
2
1
2
1
2 2
3 2
2
2
3 2
2
A
2
1 dyAt x , 0
2 dx
(As beam isloaded symmetrically)
wlC
16
wlC
16
dy wx wlEI
dx 4 16
wx wlEly x C
12 16
At x 0, y 0
C 0
wx wlEly x
12 16
x 0
wlslope
16EI
At x l / 2
wlDefletion yc
48EI
4TH SEM CIVIL SUB – ANALYSIS OF STRUCTURE CET -401
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Q.5. Find the reaction at the propped end of the cantilever as shown in figure.
2015 , 3(b)
Ans:
Downward deflection of the cantilever due to fore ‘P’
Since both the deflection are equal.
Shear force dia
Shear force at B.
Let M be the pt at a distance x from B, where shear force charges sign.
This the shear force is zero at a distance 31/8 from B
Bending moment dia.
Bending moment will be maximum at M., Where shear force changes sign
B
PL
A
B
PL
A
W /unit run
3
a
Ply
3EI
3 4Pl Wl
3EI 8EI
3Wl 3WP
8 8
B
A
3wlF
8
5wlF
8
x 35x 31 3x
1 x 5
31x
8
B
2 2
A
M 0
3wl wl wlM .1
8 2 8
4TH SEM CIVIL SUB – ANALYSIS OF STRUCTURE CET -401
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Bending moment at any section x at a distance x from the propped end B.
Find out the pt. of contraflexure, let up equate this bending moment to zero
Q. 6. Derive expression for slope and deflection of a cantilever carrying point
load at its free end using moment area method 2015 7(b)
Ans:
Let l be the length of cantilever. Let the point load W applied at B.
Let the slope at B be b.
Area of bending moment diagram
Let the deflection of B with respect to A be yb.
2 2
m
3wl 31 w 31 9wlM
8 8 2 8 128
2
m
3wl waM .x
8 2
23wl wx.x 0
8 2
31x
4
b
Area of bending moment diagram between A & B
EI
2
2
b
1 w,wl
2 2
w
2EI
b
3 3
b
Axy
EI
2x
3
w 2 wy .
2EI 3 3EI
4TH SEM CIVIL SUB – ANALYSIS OF STRUCTURE CET -401
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7 MARKS
Q.7. Calculate maximum slope and deflection in case of a cantilever of span 6m
subjected to a point load of 10 kN at free end and UDL of 10 kN/m over half
span from fixed end. Take EI constant. 2013 2(b), 2016 2(c)
Ans:
Given load at free end (W) = 10KN = 10 × 103 N. length = AB(l) = 6m = 6 × 103 mm,
Udl AC(w) = 10 kN/m = 10 N/mm lngth AC(l1) = 3m = 3 × 103 mm. E = 200 GPa =
200 × 103 N/mm2, I = 100 × 106 mm4.
Slope at the free end :
2 2 3
B
23 3
3 6
33 3 33
3 6 3 6
Wl Wl w(l l)i
2EI 6EI 6EI
10 10 6 10
2 200 10 100 10
10 6 10 3 1010 6 10
6 200 10 100 10 6 200 10 100 10
0.009 (0.018 1.75) 1.7
3 4
B
4 3
1 1
3 43 3 3
3 6 3 6
43 3
3 6
77
Deflectionat the free end :
wl wly
3EI 8EI
W l l W l l l
8EI 6EI
10 10 6 10 10 6 10
3 200 10 100 10 8 200 10 100 10
10 6 10 3 10
8 200 10 100 10
33 3 3
3 3
10 6 10 3 10 6 10
6 200 10 100 10
36 81 (75.93 94500)
117 (94575.93) 94458.93
4TH SEM CIVIL SUB – ANALYSIS OF STRUCTURE CET -401
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Q.8. Find out the prop. Reactional of a cantilever beam of span 6m subjected to
an UDL 10 kN/m2 over whole span. The beam is propped at free end.
2013,2(c)
Q.9. Write down the assumption in slope deflection method. (2013 )
Ans: Consider a bam AV subjected to a bending moment. As a result of loading let the
beam deflect from ACB to ADB into a circular arc, as shown in figure below.
Let L = Length of the beam AB.
M = Bending moment.
R = Radius of curvature of the bent up beam
I = Moment of inertia of the beam section.
E =Modulus of elasticity of beam material.
Y = Deflection of the beam.
P = Slope of the beam.
From the geometry of a circle, we know that
Given : length 6m, load10 KN/m
We know that proportion reaction.
3wl 3 10 6 180P 22.5KN
8 8 8
B
6m
A
10kN /m
22
2
AC CB EC CD
l l(2R y) y
2 2
l2 Ry y 2Ry
4
ly
8R
M Ewe know that
I R
EIR
M
4TH SEM CIVIL SUB – ANALYSIS OF STRUCTURE CET -401
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Now substituting this value of R in equation (i)
From the geometry of the figure, we find that the slope of the beam I at A or B is also
equal to angle AOC.
Since the angle i is very small, therefore sin i may be taken equal to
Again substituting the value of R in equation (ii)
Q.1. Derive an expression for the slope and deflection of a simply supported
beam subjected to a UDL of w/unit length. 2014 2(c)
Ans: Simply supported beam with uniformly distributed load
AC Lsin i
OA 2R
2 2l mly
EI 8EI8
M
li radians
2R
l l mli radians
EI2R 2EI2
M
2
x
2 2 2
2
wl wxM x
2 2
d y wl wx wlx wxEI x
dx 2 2 2 2
2 3
1
3 3
1
3
1
2 3 3
dy wlx wxEI C
dx 4 6
1 dyx , 0 As beam is loaded symmetrically
2 dx
wl wl0 C
16 48
wlC
24
dy wlx wx wlEI
dx 4 6 24
4TH SEM CIVIL SUB – ANALYSIS OF STRUCTURE CET -401
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Q.2. Derive the slope and deflection of a cantilever beam with a point load at its
free end by double integration method. 2014 6(c)
Ans: Deflection curve
Cantilever with a load at the free end
Substituting in a, C1 = 0
Substituting in b, C2 = 0
x
2
2
2
1
2 2
1 2
M W(1 x)
d yEI [ W(1 x)] Wl Wx
dx
dy WxEI Wlx C ............(a)
dx 2
Wlx WxEly Wlx C x C ................(b)
2 6
at x 0
dy0 and y 0
dx
2
2 3
2 22
B
3 3
B
3
dy WxEI Wlx
dx 2
Wlx Wxand Ely
2 6
at x 1
1 Wl WlSlope Wl
EI 2 2EI
1 Wl WlDeflection, y
EI 2 6
Wl
3EI
4TH SEM CIVIL SUB – ANALYSIS OF STRUCTURE CET -401
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Q.3. Derive the expression for maximum slope and maximum deflection in case
of a simply supported beam of span 6m subjected to an electrically placed
point load 30 kN at distance 2m and 4m respectively from supports. Take EI
constant. 2013 (4)
Ans:
Given Span l = 6m = 6 × 103 mm
Point load (wt) = 30 kN = 30 × 103 N
A = 2m = 2 × 103 mm, b = 4mm = 4 × 103 mm
EI = 26 × 1012 N-mm2
Slope at A
2 2
A
3 32
3 2 3
12 3
6 6
8
6
8
2 2
B
3 3
wbi (l b )
6EIL
(30 10 ) (4 10 )(6 10 ) 4 10
6 (26 10 )(6 10 )
12(36 10 ) (16 10 )
6 (26 10 ) 6
12(20 10 )
36 (26 10 )
12 20.0025 rad
36 (260)
Slope at B
wai (l a )
6EIL
(30 10 ) (2 10 )
6 (2
12 3
3 2 3 2
6 6
8
6
8
6 10 )(6 10 )
(6 10 ) (2 10 )
6(36 10 ) (4 10 )
6 (26 10 ) 6
6(32 10 )
36 (26 10 )
4TH SEM CIVIL SUB – ANALYSIS OF STRUCTURE CET -401
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Q.4. Derive an expression for slope and deflection of a simply supported beam
subjected to a central point load. 2015 1(c)
Ans: Let us consider a simply supported beam subjected to a central point load.
In order to attain a single expression for bending moment which will apply across the
complete beam in this case it is convenient to take the origin for x at the centre, then :
2
2 2 2
C
3 3 3
12 3
3 2 3 2 3
5
6 6 6
5
6 2 1920.00205
36 26 10 93600
wabDeflection (l a b )
6EIL
(30 10 ) (2 10 )(4 10
6 (26 10 )(6 10 )
(6 10 ) (2 10 ) (4 10 )
3 2 4
6 (26 10 ) 6
(36 10 ) (4 10 ) (16 10 )
24(32 10
36 (26 10 )
6 6
6
5
) (20 10 )
24(16 10 )
36 (26 10 )
24 160 38404.102mm
36 26 936
2
xx 2
2
2 2
d y w L WL WxM EI x
dx 2 2 4 2
dy WL WxEI x A
dx 4 4
{Integrating both side)
WLx WxEIy Ax B
8 12
{Again by int egrating}
3 3
3 3 3
2 3 3
3
max
2
max
dyAt x 0, 0, A 0
dx
L WL WLx my0, 0
2 32 48
WL WL WLB
96 32 48
1 WLx Wx WLy
EI 8 12 48
WLSlope, y at the centre
48EI
dy WLDeflection, at the ends.
dx 16EI
4TH SEM CIVIL SUB – ANALYSIS OF STRUCTURE CET -401
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Q.Find slope and deflection for a s/s beam with a UDL over the span by
double integration method 7(c) 2017
Consider a secn x-x at a distance x from the end A
2 2
2
2 3
2 2
2 2
int
12 2 2 3
wl xMx X x w X x X
d Y wlx wxEI
dx
egrating we get
dy wl x w xEI c
dx
4TH SEM CIVIL SUB – ANALYSIS OF STRUCTURE CET -401
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1. ., , 0
2,
3 30 1
16 48
3 3 3 3 31
16 48 48
31
24
2 3 3
4 6 24
dyi e x
dx
wl wlC
wl wl wl wlC
wlC
dy x wx wlEI wl
dx
Slope at “A”
Putting X=0, we 𝑊𝑙3
24
𝜃A = −𝑤𝑙3
24𝐸𝐼
Integrating the slope eq1 we get
3 4 3
3
4
24 3 6 4 24
0, 0, 2 0
3 4 3
12 24 24
2
4 3max
12 8 24 16 24 2
5 4max
384
5
384
n n
n
WL x w x wlEUY x c
when x y C
wlx wx wl xEIy Defl en
lat mid span x
Wl x l wl wl x lwe getEI y
x x x
wlY
EI
wlDownward defl Yman
EI
4TH SEM CIVIL SUB – ANALYSIS OF STRUCTURE CET -401
25
CHAPTER:3
2 MARKS
Q.1. State any two advantages to fixed beam. 2015 , 3(a)
Ans: i. The fixed beam is subjected to a lesser maximum bending moment than the
simply supported beam carrying the same, load.
ii. For the same loading maximum deflection a fixed beam is less than that if the
simply supported beam.
Q.2. Write down the expression for fixed and moment for a fixed beam of span ‘l’
subjected to a point load ‘w’ at distance ‘a’ and ‘b’ from both ends
2013 (s) (New) 1(f)
Ans: The expressions are
2
2
2
2
wabMa
l
wa bMb
l
4TH SEM CIVIL SUB – ANALYSIS OF STRUCTURE CET -401
26
5 MARKS
Q.1. A fixed beam AB of span 6 m is subjected to two point load of 20 kN and 30
kN at a distance of 2m and 4 m from A. Calculate fixing moment at A and B.
2015 2(b)
Ans:
Q.2. A fixed beam AB of span 6m is subjected to a UDL of 2 kN/m. Determine the
S.F. & B.M.D., its diagram 2014, 3(b)
Ans:
Consider any section ‘X’ at a distance x from the left end ‘A’. The shear force
at the section is :
2
2
2 2
2 2
2
2
2 2
2 2
WabFixing moment at A
20 2 4 30 2 231.12kN m.
6 6
Wa bFixing moment at B
20 2 4 30 4 235.56kN m.
6 6
A B
WL 2 6R R 6kN
2 2
x A
A A
B A
f R w.x 6 2.x.
wxAt A, x 0, hence F R
2
2 66 6kN( )
2
At B,x 6, hence F R w.x
6 2 6 6kN.
4TH SEM CIVIL SUB – ANALYSIS OF STRUCTURE CET -401
27
Simply supported bending moment at the section ‘X’ at a distance ‘x’ from left end ‘A’ is given by
7 MARKS
c A
2
2
C
2
2 2
L 6At C, x 3
2 2
Hence f R w.x
6 2 3 0.
26 6 6 0
2
w L w LAt, C, x ,hence M . .
2 2 2 2 2
6 2 66 18 9 9 kN m( )
2 2 2
Fixed end BM at A and B
Wl 2 6( ) 6kNm
12 12
x A
22
A
2
B
xM R w.x.
2
2x6x 6x x
2
The value of B.M. at different po int s are :
w w.0At A, x 0, hence M 0 0
2 2
At B, x L 6
w wHence M .L L
2 2
4TH SEM CIVIL SUB – ANALYSIS OF STRUCTURE CET -401
28
Q.1. A fixed beam is subjected to an UDL over whole span. Derive the expression
for fixed end moments. 2013, 2(d) 2016, 4(c)
Ans:
Let mA = Fixing moment at A and
MB = Fixing moment at B
Since the beam is symmetrical therefore mA and mB will also be equal.
Now equating the areas of the two diagrams.
We know that maximum positive bending moment at the centre of the beam = wl2/8.
Net positive bending moment at the centre of the beam
Shear force diagram, Let RA = Reaction at A, and RB = Reaction at B.
Equating the clockwise moments and anticlockwise moments about A.
Deflection of the beam,
We know that bending moment at any section X, at a distance x from A,
2 3
A
2
A
2
B
2 wl wlm l l.
3 8 12
wlm
12
wlSimilarly, m
12
2 2 2wl wl wl
8 12 24
B 0 B
2
B
A
lR l m m w l
2
wlR
8
wlSimilarly, R
2
4TH SEM CIVIL SUB – ANALYSIS OF STRUCTURE CET -401
29
Integrating the above equation
Where C1 is the first constant of integration. We know that when x = 0, the dy/dx = 0
Therefore C1 = 0
Integrating the equation (ii) once again
Where C2 is the second constant of integration, We know that when x = 0, then y = 0
Therefore C2 = 0
We know that the maximum deflection occurs at the centre of the beam. Therefore
substituting x = l/2 in the above equation.
Point of contraflexures.
The points of contrflexures may be found out by equation (i) to zero.
2 2'
x x x
2 2 2
2
wl wx wlM x
2 2 12
d y wlx wx wlEI ...................(i)
dx 2 2 12
2 3 2
1
dy wlx wx wl xEI C
dx 4 6 12
2 3 2dy wlx wx wl xEI .................(ii)
dx 4 6 12
2 3
2
dy wlx wxEI _ C
dx 4 6
3 4 2 2wlx wx wl xEI.y ..............(iii)
12 24 24
3 4 22
c
4 4 4 4
4
c
wl l w l wl lEI.y
12 2 24 2 24 2
wl wl wl wl
96 384 96 384
wly
384EI
4TH SEM CIVIL SUB – ANALYSIS OF STRUCTURE CET -401
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Solving this quadratic equation for x,
CHAPTER:4
2 MARKS
Q.1. Explain theorem of three moment, 2014, 5(a), 2015, 5(a)
Ans: Theorem of three moments: If a beam has a support, the end ones being fixed, then
the same number of equations required to determine the support moments may be
obtained from the consecutive pairs of spans.
Theorem of three moments states that
Where MA, MB, MC = Support moments at A, B and C.
2 2
22
22
wlx wx wl0
2 2 12
llx x 0
6
lx lx 0
6
22 4l
l l6x
2
l l
2 2 3
0.5l 0.289l 0.789 l and 0.211 l.
1 1 2 2A B c
1 1 1 2
1 1 2 2
1 1 2 2
l l l lM 2M M
I I I I
A x A x6
I l I l
4TH SEM CIVIL SUB – ANALYSIS OF STRUCTURE CET -401
31
A1, A2 = Area of B.M. diagram for the given loading (s/s)
X1,x2 = Distance of centroid of areas A1 and A2 from A and C respectively.
l1,l2 = Span AB and BC respectively
I1, I2 = MI. of span AB and BC respectively.
Q.2. Write down the expression for three moment equation with usual meaning
2013, 2(i)
Ans: Three moment equation:-
6 MARKS
Q.1. A continuous beam is simply supported over two spans, such that AB = 6m
and BC = 4m. It carries uniformly distributed load of 2 kN/m over span BC
and a point load of 5 kN at the centre of span AB. Determine the support
moment over B by applying theorem of three moments. 2014,2016
Ans:
1 1 2 2A 1 B 1 2 0 2
1 2
6a x 6a xM l 2M (l l ) M l
l l
AB
C
5kN 2kN
1 1 2 2A 1 B 1 2 C 2
2 2
D
6a x 6a xM l 2M (l l ) M l
l l
wl 5 6M 7.5 kNm
4 4
4TH SEM CIVIL SUB – ANALYSIS OF STRUCTURE CET -401
32
Bending moment at the mid of the span BC
Now using three moment equation.
Q.2. A continuous beam ABC is simply supported over support ‘A’, ‘B’ and ‘C’. the
span AB is 5m and BC is 6m. It is subjected to a point load 30 kN at mid
span of AB and an UDL 10 kN/m over whole span of BC. Find out the
moment at support ‘B’. The beam is of uniform cross-section.2013 2(c)
Ans:
Let mA = Fixing moment at A.
MB = Fixing moment at B.
MC = Fixing moment at C.
2 2
1 1
2 2
wl 2 44kNm
8 8
1 2 1 3a x 3 7.5 3 3 7.5 3
2 3 2 3
22.5 45 67.5
2a x 4 4 2 21.33
3
1 1 2 2A 1 B 1 2 C 2
1 2
B
B
B
6a x 6a xM l 2M (l l ) M l
l l
6 67.5 6 21.330 2M (6 4) 0
6 4
20M (67.5 31.99) 99.495
M 4.97 Nm
D
2 2
2
1 1
wl 30 5m 37.5 KN.m
4 4
bending moment at the mid of the span BC
wl 10 6 10 36 36045.00 KN.m
8 8 8 8
We find that
1 2 2.5 1 2.5a x 2.5 37.5 2.5 37.5 2.5
2 3 2 3
78.12 (46.87 3.33)
78.12 156.07 23
2 2
4.19
2a x 45 6 3 540
3
4TH SEM CIVIL SUB – ANALYSIS OF STRUCTURE CET -401
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Now using three moments equation
Shear force diagram
Let RA = Reaction at A
RB = Reaction at B
RC = Reaction at C
Taking moments about B.
7 MARKS
Q.1. A continuous beam ABC 10m long rests on three support A, B and C at the
same level and is loaded as shown in figure.
Determine the moments over the beam and draw the bending moment
diagram. Also calculate the reactions at the support and draw shear force
diagram using theorem of three moment.
Ans: Applying the theorem of three moments of the span AB and AC
1 1 2 2A 1 b 1 2 c 2
1 2
b
B
B
6a x 6a xm l 2m l l m l
l l
6 234.19 6 5400 2m (5 6) 0
5 6
22m (281.02 540) 821.02
821.02m 37.31
22
B A A
A
A
C
C
C
B
R R 5 30 2.5 37 /31 R 5 75
R 5 37.31 75
37.31 75 37.69R 7.53kN
5 5
SimilarlyR 6 (60 3) 37.31
R 6 37.31 (60 3)
37.31 (60 3)R 23.78
6
R (30 10 6) (7.53 23.78) 90 31.31 58.69
4TH SEM CIVIL SUB – ANALYSIS OF STRUCTURE CET -401
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Length of AB = l1 = 6 m
Length of BC = l2 = 4m
MA = Fixing moment of A
MB = Fixing moment of B
MC = Fixing moment of C.
Let us consider the beam AB as simply supported so bending moment at D.
Bending moment at the mid of the span BC.
Let RA = Reaction at A
RB = Reaction at B
RC = Reaction at C
Taking moment about B.
MB = RA × 6 – 3 × 4
4.002 = RA × 6 – 12
RA = {4.002 + 12} × 1/6 = 2.667 kN
Again MB = RC × 4 – 4 × 2
4.002 × RC × 4 – 8
RC = (4.002 + 8) × ¼ = 3.0005 kN.
D
1
wab 3 2 4M 4 kN m
l 6
2 22
1 1 2 2A 1 B 1 2 C 2
1 2
Wl 3 46 kN m.
8 8
Using three moment equation.
6a x 6a xM l 2M (l l ) M l
l l
2
1
1
2
B 1 2
A C
B
1a 6 4 12m
2
6 2 8x 2.67m.
3 3
x 2m.
6 12 2.67 6 16 22M (l l )
6 4
80.04 M M 0 kN m
80.04M 4.002 kN m.
2 (6 4)
4TH SEM CIVIL SUB – ANALYSIS OF STRUCTURE CET -401
35
fy=0
RA + RB + RC = 3 + 4
2.667 + RB + 3.0005 = 7
RB = 7 – (2.667 + 3.0005)
RB = 1.3325 kN.
Q.2. A continuous beam ABCD is simply supported over three spans, such that AB
= 6m. BC = 8 m and CD = 5m. It carries UDL of 4 kN/m in span AB, 3 kN/m
in span BC and 2 kN/m in span CD. Find the support moments B and C and
draw the SF and BM diagrams. 2014 4(c)
Ans:
Applying the theorem of three moments for the spans AB and BC.
Since A is the simply supported end of the girder.
Ma = 0
28MB + 8MC = 216 + 384 = 600
14MB + 4MC = 300 ……………………(1)
Consider the span BC and CD.
Applying the theorem of three moments for these spans.
Since D is the simply supported end.
Md = 0
8Mb + 26MC = 384 + 52.5 = 446.5
4Mb + 13MC = 223.25 ……………………..(2)
14Mb + 4Mc = 300
4Mb + 13Mc = 223.25
a B C
3 3
M 6 2M (6 8) M 8m
4 6 3 8
4 4
3 3
B C d
3 8 2 5M 8 2M (8 5) M 5
4 4
4TH SEM CIVIL SUB – ANALYSIS OF STRUCTURE CET -401
36
10Mb – 9Mc = 76.75
10 Mb –76.75 + 9Mc
4Mb + 150 .79 = 223.25
4Mb = 72.45
Mb = 18.11 kN –m
Maximum free bending moment for span AB.
Maximum free bending moment for span CD
B.M. at c = 15.01 × 14 + Vb × 8 – 4 × 6 × 11 – 3 × 8 × 4 = 11.599
Vb = 11.599 – 210.14 + 264 + 96
= 11.599 – 210.14 + 360
Vb = 161.459 kN
Vd = 2.68kN
Vc = total load – (Va + Vb + Vd)
= (4 × 6 + 3 × 8 + 2 × 5) –(15.01+161.459+2.68)
= (24 +24 + 10) – 179.149 = 58 – 179.149 = –121.149
CB
CC
76.75 9MM
10
76.75 9M4 13M 223.25
10
24 618kN m
8
2
2
a
a
a
2 56.25kN m
8
4 6B.M atB V 6 18.11
2
18.11 72V
6
V 15.01kN
2
d
d
2 5B.M. at c V 5 11.599
2
V 5 11.599 25
4TH SEM CIVIL SUB – ANALYSIS OF STRUCTURE CET -401
37
Q.3. A continuous beam ABC with fixed end at ‘A’ simply supported over support
‘B’ and ‘C’. the span AB = 6m and BC = 5m. Span AB is subjected to a point load
20 kN at 2m from support ‘A’ and UDL 5 kn/m over whole span BC. Find out the
reactions and moment at supports. Draw the shear force and bending moment
diagram for the same. 2016 5(c) 2013 (s)
Ans:
Support moments at A, B and C.
Let mA = Support moment at A
MB = Support moment at B
MC = Support moment at C
Bending moment under the 20 KN load at AB
Consider the beam BC as a simply supported beam. Therefore bending moment at the
mid of span BC.
Geometry of the above bending moment diagram. We find that for the span OA and
AB.
= [142.19+(26.66) (4.67)
=(142.19 + 124.50)=266.69
Similarly for the spans AB and BC.
1
wab 20 2 426.66 KN.m
l 6
2
2wl 5 515.63 KN.m
8 8
0 0
1 1
a x 0
1 2 4 1 2a x 26.66 4 26.66 2 4
2 3 2 3
1 1
2 2
A B
A B
A B
1 2 4 1 2a x 26.66 4 26.66 2 4 266.69
2 3 2 3
2a x 15.63 5 2.5 130.25
3
12m 6m 266.69
6(2m m ) 266.69
266.692m m 44.45..........(i)
6
4TH SEM CIVIL SUB – ANALYSIS OF STRUCTURE CET -401
38
Now using three moments equation for the span AB and BC.
6mA + 22mB = –[266.69+156.30]
6mA + 22mB = 422.99 ……………….. (ii)
Solving equation (i) and (ii)
Equation (i) × 3 = 6mA + 3mB = 133.35
Equation (ii) × 1 = 6mA + 22mB = – 422.99
Put the value of mB in equation (i)
2 mA + (–29.28) = – 44.45
2 mA – 29.28 = – 44.453
2 mA = – 44.45 + 29.28
= – 15.17
mA = – 7.58 Kn-m
mB = – 29.28 KN-m
mC = 0
Shear force diagram
Let RA = Reaction at A
RB = Reaction at B
RC = Reaction at C
Taking moment about B
–29.28 = RC × 5 – (25 × 2.5) – 29.28 = 5RC – 62.5
5 RC = –29.28 + 62.5 = 33.22
1 1 2 2A 1 B 1 2 C 2
1 2
A B
6a x 6a xm l 2m (l l ) m l
l l
6 266.69 6 130.25m 6 2m (6 5) 0
6 5
B
B
19m 556.34
556.34M 29.28
19
A
15.17m 7.58
2
C
33.22R 6.65 KN
5
4TH SEM CIVIL SUB – ANALYSIS OF STRUCTURE CET -401
39
Now taking moment about A.
– 7.58 = RB × 6 – (25 × 8.5) – (20 × 2) + 6.65 × 11
– 7.58 = 6RB – 212.5 – 40 + 73.15
6RB = –7.58 + 212.5 + 40 – 73.15 = – 80.73 + 252.50 = 171.77
CHAPTER:5
Q.1. Define carry over factor , Stiffness factor 2013 1(j) 2015 4(a)
Ans: Carry over factor: The ratio of moment produced at a joint to the moment applied
at the other joint, without displacing , it is called Carry over Factor.
Stiffness Factor: Moment required to rotate end by unit angle, when rotation is
permitted at that end, is called stiffness factor.
Q.2. state the Stiffness Factor for a beam fixed at one end & freely supported at
the other. 2014 4(a)
Ans: The stiffness factor at fixed end,
K1=4EI/L
The Stiffness Factor at J/S end
K2=3EI/L
Q.3. Find the other State the stiffness factor for a beam fixed at our end
& freely supported at. 2017 4(a)
Ans : K=4𝐸𝐼
𝑙
(one end fixed s Free support at the other)
Q.4. State the stiffness factor for a beam fixed at one end & freely
supported at the other 2017 4 (a)
Ans:- K = 4𝐸𝐼
𝐿 (𝑂𝑛𝑒 𝑒𝑛𝑑 𝐸𝑖𝑥𝑒𝑑 & 𝐹𝑟𝑒𝑒𝑙𝑦 𝑠𝑢𝑝𝑝𝑜𝑟𝑡𝑒𝑑 𝑎𝑡 𝑡ℎ𝑒 𝑜𝑡ℎ𝑒𝑟)
B
A
171.77R 28.62KN
6
R 20 2 28.62
40 28.62 11.38 KN
4TH SEM CIVIL SUB – ANALYSIS OF STRUCTURE CET -401
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7 MARKS
Q.1. A continuous beam ABC with fixed end at ‘A’ S/S over support ‘B’ and ‘C’.
the span AB = 6m & BC = 5m, Span AB is subjected to a point load 20 kN at 2m
from support ‘A’ & UDL 5 kN/m over whole Span BC. Find outthe reactions
moment of supports by suing moment distribution method.
2016 5(c) 2014 (s)
Ans:
Let us assume the continuous beam ABC to be split up into filxed beams AB, BC.
In span AB, fixing moment at A
In span BC, fixing moment at B
Now let us find out he distribution factor at B. From the geometry of the figure
we find that the stiffness factor for BA.
2 2
2 2
2 2
2
wab 20 2 (4)17.77kNm
l (6)
Fixing moment at B,
wab 20 2 48.88 kNm
l 6
2 2
2
wl 5 510.41kNm
12 12
wlFixing moment of c 10.41kNm
12
BA
BC
4EI 4E I 2k EI
l 6 3
4EI 4E I 4k EI
l 5 5
Distribution factor for BA and BC
2 4EI EI
5EI 6EI3 5and and2 4 2 4 11 11
EI EI EI EI3 5 3 5
–17.77 8.88
5/11
–10.41 10.41
6/11
Fixed end moment
–10.41
-5.41
Release c.
Carry over
-17.77 8.88
3.06
1.53
0 0
-15.62 0
3.67
0 0
Initial moments
Distribute
Carryover istribute
Final moment –11.95 0 –16.24 11.94
4TH SEM CIVIL SUB – ANALYSIS OF STRUCTURE CET -401
41
8 MARKS
Q.1. Analyse the portal frame shown in figure using moment distribution method
and draw the bending moment diagram 2015, 7(c)
Ans:
FAB FBA CD FDC
2
FBC
FCB
Fixed End Moment
M M M M 0
W 2 6 6M 6 kNm
12 12
M 6 kNm.
4TH SEM CIVIL SUB – ANALYSIS OF STRUCTURE CET -401
42
Free moment diagram of BC parabola with maximum ordinate
Draw in compression side of member. Difference of these two diagram is BM
diagram. Positive and negative are marked upon whether cousing tension/compression on
doted side.
Q.2. A rectangular portal frame of uniform flexural rigidity EI, carries a UDL of 20
kN/m as shown in fig. Draw the bending moment diagram and sketch the
deflected curve. If L = 4 m and EIAB = EIBC = EICD.
2014(s)
Ans: Given data
Length of AB =4m
Length of Bc = 2 L = 2 × 4 = 8 m2
Length of CD = 4m
Load on BC (A) = 20 kN/m
EIAB = EIBC = EICD
Support Reactions:
Let MAB = Moment at Aa
MBA = Moment at B in span BA.
MBC = Moment at B in span BC.
MCB = Moment at C in span CB.
2 6118kNm
8
4TH SEM CIVIL SUB – ANALYSIS OF STRUCTURE CET -401
43
MCB = Moment at C in span CD.
MDC = Moment at D.
(i) Fixed end moments: Let us assume the frame to be made up of fixed beams AB, BC
and CD from the geometry of the figure, we find that fixed end moment at A.
(ii) Slope Deflection Equation : Since frame is fixed at A and D. therefore the slopes iA
and iD will be equal to zero.
Moment at ‘A’ in the span AB,
' '
AB BA
2 2'
BC
2 2'
CB
' '
CB DC
M M 0
wL 20 8M 106.67kN m
12 12
wL 20 8M 106.67kN m
12 12
M M 0
'ABAB A B AB
B
B
'BABA B A BA
'
B A BA
B
B
2EIM 2i i M
2EI(0 i ) 0
4
EI i
2
2EIM 2i i M
2EI2i i M
4
2EI(2i 0) 0
4
EI i
Moment at 'B ' in span BC,
'BCBC B C BC
B C
B C
'CBCB C B CB
C B
C B
2E IM (2i i ) M
2I I(2i i ) 106.67
63
EI(2i i )106.67
3
2E IM (2i i ) M
2EI(2i i ) 106.67
82
EI(2i i )106.67
2
4TH SEM CIVIL SUB – ANALYSIS OF STRUCTURE CET -401
44
Moment at ‘C’ in span CD.
'CDCD C D CD
C
C
'DCDC D C DC
C
C
2E IM (2i i ) M
2EI(2i 0) 0
4
EI i
2EIM (2i i ) M
2EI(0 i ) 0
4
EI i
2
4TH SEM CIVIL SUB – ANALYSIS OF STRUCTURE CET -401
45
4TH SEM CIVIL SUB – ANALYSIS OF STRUCTURE CET -401
46
4TH SEM CIVIL SUB – ANALYSIS OF STRUCTURE CET -401
47
Q.3. Analyze a symmetrical rectangular portal frame using deflection method of
horizontal span 5m subjected to an UDL 10 kN/m over whole span and
height of 3.5 m. 2013, (7)
Ans:
The ends A and D of the frame ABCD are fixed and therefore
A = D = 0
As the portal frame is symmetrical and loaded symmetrically rotation 3 = c
and there will be no way i.e. = 0.
The fixed moments are :
The slope deflection equations in terms of unknown are :
For equilibrium the sum of the moments at joint B is zero.
BC
CB
10 5 5M 20.83kNm
12
M 20.83kNm
BAB B
BBA B
BC 3 B
B
2EI2EIM ( 0)
3.5 3.5
4EI2EIM ( 2 )
3.5 3.5
2EIM (2 ) 20.83
5
2EI 20.83
5
4TH SEM CIVIL SUB – ANALYSIS OF STRUCTURE CET -401
48
iii) Equilibrium equations: Since the joints B and C are in equilibrium, so first equation
MBA + MBC equal to zero.
(EI × iB) + [EI (2 iB + iC) – 106.67] = 0
(EI × iB) + (2EI × iB) + (EI × iC) – 106.67 = 0
3 (EI × iB) + (EI × iC) = 106.67.
Now equation MCB + MCD = 0
[EI(2iC + iB) + 106.67] + (EI × iC) = 0
2(EI × iC) + (EI × iB) + 106.67 + (EI × iC) = 0
3(EI × iC) + (EI × iB) + 106.67 = 0
By symmetry, iB = iC, substituting these values, we get 2EI × iB = 106.67
AB BC
BB
B
B
B
M M 0
2EI 2EI 20.83 0
3.5 5
1 12EI 20.83 0
3.5 5
2EI 0.48 20.83
21.69
EI
BAB
BBA
BC B
CB B
2EI 2EI 21.69M 12.39kNm
3.5 3.5 EI
4EI 2EI 21.69M 24.78kNm
3.5 3.5 EI
2M EI 20.83
5
2 21.69EI 20.83 12.15kNm
5 EI
2 2 21.69M EI 20.83 EI 20.83
5 5 EI
12.15kNm
B
C
C
106.67EI i 53.335
2
2EI i 106.67
106.67EI i 53.335
2
4TH SEM CIVIL SUB – ANALYSIS OF STRUCTURE CET -401
49
Final moments :
The bending moment at the mid of the span BC, buy considering it as a simply
supported beam.
CHAPTER:6
2 MARKS
Q.1. In case of a column whose both ends are hinged , what will be its equivalent
length. 2015, 4(a)
Ans: In Case of a column whose both ends are hinged. Then its equivalent length will be
same as its actual length le = l
BAB
BA B
B CBC
B C
C BCB
C B
EI iM
2
53.33526.667 kN m.
2
M EI i 53.335 kN m.
EI(i i )M 106.67
3
(2EIi ) (EIi )106.67
3
106.67 ( 53.335)106.67
3
88.89 kN m.
EI(2i i )M 106.67
2
2EIi EIi106.67
3
106.67 53.335106.67
2
80.0
0 kN m.CD C
CDC
M EI i
53.335 kN m.
EI iM
2
53.33526.667 kN m.
2
2 2w 20 8160 kN m.
8 8
4TH SEM CIVIL SUB – ANALYSIS OF STRUCTURE CET -401
50
5 MARKS
Q.1. A steel rod 5m long and 40 mm dia. is used as a column with one end fixed
& other free. Determine the cripping load by Euler’s formula. Take l as 200
GPa. 2016, 7(b)
Ans: Given data l = 5 m = 5 × 103 mm
D = 40 mm
L = 200 GPa = 200 × 103 N/mm2
One end fixed other free le = 2l
Bu Euler’s formula
Q.2. State different end conditions of column and write down the relation
between equivalent length and actual length in each case. 2015, 6(b)
Ans: In actual practice there are a number of end conditions, for columns. But we shall
study the Euler’s column theory on the following four types of end conditions, which
are important from the subject point of view.
2
Eulers 2
4 44
2 3
Eulers 3 2
EIP
le
d (40)I 125663.7mm
64 64
200 10 125663.7P
(2 5 10 )
24805N
24.85 N
Types of End Conditions Relation between equivalent
Length (Le) & actual length
(ℓ)
1. Both ends hinged
2. One end fixed and the
other free
3. Both ends fixed
4. One end fixed and the
other hinged
Le = ℓ
Le = 2ℓ
Le = ℓ/2
Le = ℓ/√2
4TH SEM CIVIL SUB – ANALYSIS OF STRUCTURE CET -401
51
Q3.- State different end conditions of calumn and write down the
relation between equivalent length , actual length in each case.
20176(b)
Ans
End conditions
1) Both ends hinged
2) One end fixed and the other free
3) Both ends fixed
4) One end fixed and the other hinged
Relation between equivalent length (Le) and actual length (L)
Le=l
Le=2l
Le=𝑙
2
L2=𝑙
√2
CHAPTER:7
2 MARKS
Q.1. What do you mean by three hinged arch ? 2013, 1(d), 2016, 6(a)
Ans: Vector diagram L Diagram showing the magnitude of forces along with direction is
called vector dig. Polar diagram: Diagram showing magnitude of forces is called polar
diagram.
5 MARKS
4TH SEM CIVIL SUB – ANALYSIS OF STRUCTURE CET -401
52
Q.1. A three hinged symmetrical arch of span 7m is subjected on UDL of w over
whole span. Draw the S.F. & B.M.D. for the arch 2013, 2(f)
Ans:
Let the rise of the arch = hm , Due to symmetry
VA = VB = ½ × total load.
CHAPTER-7
5 MARKS
Q.1. A three-hinged parabolic arch of span 40m, and rise 10 m is carrying a UDL
as shown in the fig. Find the horizontal thrust at the springing. 2014,5(b)
A
2
2
2
A
2
2 2
2
2
1 w 7w.L 3.5w
2 2
Taking moment about c, we get
1 L LO V . Hh w .
2 2 4
WL L wLHh
2 2 8
wL 49wH
8 8
At any sec tion dis tance x from A
wLM V x Hy
2
4hx(L x)But in parabolic arxh y
L
wL wL 4hx(L x) wLM x .
2 8 L 2
wLx w wLx(L x)
2 2
02
4TH SEM CIVIL SUB – ANALYSIS OF STRUCTURE CET -401
53
Ans: Given data,
Span (l) = 40m and central rise (yC) = 10m.
H = Horizontal thrust at the springing.
VA = Vertical reaction at A.
VB = Vertical reaction at B.
Vertical reaction VB at B can be calculated by taking moments about A and equating
anti-clockwise moments with clockwise moments.
The beam moment at C due to external loading
c = VB × 20 = 150 × 20 = 3000 kN-m.
Horizontal thrust,
Q.2.
Find out the reaction at A and B and draw the bending moment diagram for
the parabolic arch as shown in figure. 2015, 6(c)
B
B
V 40 (30 20) 10 6000
6000V 150kN.
40
c
c
3000H 300kN
y 10
4TH SEM CIVIL SUB – ANALYSIS OF STRUCTURE CET -401
54
Ans:
Bending moment diagram is a triangle with maximum ordinate at load point.
At mid-span the net bending moment is zero. Ordinate of the beam moment diagram
is zero. Ordinate of bending moment diagram at mid span is :
Parabola drawn with central ordinate equal to 60 kNm.
B
A
A
A
A B
B
M 0
V 20 200(20 6) 0
V 20 2800 0
2800V 140kN
20
V V 200kN
V 200 140 60kN
200 6(20 6) 200 6 1484kNm
20 20
c
84 1060kN
14
M 0(In Arch),Hh 60kN
60 60H 12kN.
h 5
BM Bending Moment 2
4hx(L x)Y Y Hy H
4TH SEM CIVIL SUB – ANALYSIS OF STRUCTURE CET -401
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