error free detailed solution · (200 meter from riddhi siddhi tiraha), gopal pura mode (between...
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ENGINEERS PRIDE Class Room/Office Address-C-225, Ganesh Marg, C-Block, Mahesh Nagar
(200 Meter from Riddhi Siddhi Tiraha), Gopal Pura Mode (between Gandhi Nagar Railway Station
and Durga Pura Railway Station), Jaipur, Rajasthan, 9660807149, 7014320833, 8448449932
P a g e 1 | 27
ERROR FREE DETAILED SOLUTION
RPSC AE MAINS 2018 CE P1(TECH) FULL SYLLABUS
ENGINEERS PRIDE TEST DATE 7TH NOV 2019 (SET B)
PART-A
Marks: 40
Note- Attempt all the twenty questions. Each question
carries 2 marks. Answer should not exceed 15 words
Q.1 If principal stresses in a two-dimensional case are −10Mpa and 20Mpa respectively, then
maximum shear stress at the point is
Solution:
21 2max
10 2015 N/mm
2 2
− − −= = =
Q.2 A cantilever beam of span L is loaded with a concentrated load P at the free end. Find the
deflection at the free end?
Solution:
P
A B
L
ENGINEERS PRIDE Class Room/Office Address-C-225, Ganesh Marg, C-Block, Mahesh Nagar
(200 Meter from Riddhi Siddhi Tiraha), Gopal Pura Mode (between Gandhi Nagar Railway Station
and Durga Pura Railway Station), Jaipur, Rajasthan, 9660807149, 7014320833, 8448449932
P a g e 2 | 27
Deflection at B=3
3B
PL
EI =
Q.3 What is the modulus of resilience?
Solution:
Modulus of resilience:
Maximum strain energy per unit volume which can be stored in a material up to elastic
limit is known as modulus of resilience.
I.e. proof resilience per unit volume is termed as modulus of resilience.
Proof resilience
MORVolume
=
1 1
2 2
P l
A l
= =
21
2 2E
= =
Q.4 A solid shaft having diameter 80 mm is subjected to a torque of 6kN-m. Calculate
maximum shear stress developed?
Solution:
( )
62
4
. 6 1059.68 N/mm
8032
T r
J
= = =
Q.5 Define void ratio and porosity?
Solution:
Void ratio; v
s
Ve
V= and Porosity; vV
V =
Where;
ENGINEERS PRIDE Class Room/Office Address-C-225, Ganesh Marg, C-Block, Mahesh Nagar
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P a g e 3 | 27
vV →Volume of void
sV →Volume of solid
V →Total volume i.e. v sV V+
Q.6 Write the equation for the zero air void line.
Solution:
Equation for the zero air voids line 1
wd
G
wG
=
+;
Where;
G → Specific gravity of soil
w →Unit wt. of water
w→Water content of soil
Q.7 Define coefficient of compressibility of soil.
Solution:
Coefficient of compressibility is defined as the ratio of change in void ratio to change in
effective stress.
Coefficient of compressibility v
ea
−=
Note- Coefficient of compressibility is always –ve as the void ratio decreases with increase
in stress.
Q.8 For a saturated sand deposit, the void ratio and the sp. gravity of solids are 0.70 and 2.67
respectively. Find the critical hydraulic gradient.
Solution:
Critical hydraulic gradient 1 2.67 1
0.981 1 0.7
c
Gi
e
− −= = =
+ +
Q.9 What is static indeterminacy?
Solution:
ENGINEERS PRIDE Class Room/Office Address-C-225, Ganesh Marg, C-Block, Mahesh Nagar
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P a g e 4 | 27
Static indeterminacy of a structure is defined as
static equilibrium
No.of unknown forces equations available
.s
No ofD
= −
Q.10 Write the method of analysis of determinate truss.
Solution:
There are two method of analysis of determinate truss.
1. Method of joint
2. Method of section
Q.11 What is stiffness of a member?
Solution:
The load required to produce a unit deflection is called stiffness (K).
P MK
= =
Q.12 What are the parameters include for partial safety factor used for material strength?
Solution:
Partial safety factors are used for material strength allowing uncertainties of element
behaviour, possible strength reduction due to manufacturing or costing tolerance and
imperfections in the materials.
Q.13 The flexural strength of M30 concrete as per IS 456:2000 is-
Solution:
As per IS456:2000, Modulus of elasticity= 5000 ckf
= 5000 30
=27386.127 Mpa
Q.14 What are the criteria to provide doubly reinforced beam?
Solution:
ENGINEERS PRIDE Class Room/Office Address-C-225, Ganesh Marg, C-Block, Mahesh Nagar
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P a g e 5 | 27
Following are the criteria to provide doubly reinforced beams:
• When size of beam is restricted either due to architectural point of view or headroom
consideration.
• When the beam has to take bending moment more than moment of resistance of
singly reinforced balance section.
Q.15 What is concordant profile?
Solution:
If the cable is provided in such a way that there is no change in support reaction due to
pre stressing force, it is called concordant profile. The shape of concordant profile is
similar to bending moment diagram. If concordant profile is used, secondary moments
will not be develop in the span.
Q.16 Define ‘principal rafter’?
Solution:
One of the upper diagonal members of a roof truss supporting the purlins on which the
common rafter rest is called principal rafter.
Q.17 What are the different types of tension members are used?
Solution:
ENGINEERS PRIDE Class Room/Office Address-C-225, Ganesh Marg, C-Block, Mahesh Nagar
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P a g e 6 | 27
Different types of tension members are:-
(1) Rods and bars
(2) Single structured shapes and plates
(3) Wires and cables
(4) Built up members
Q.18 What are the different modes of failure of column?
Solution:
Following are the modes of failure of column.
• Crushing
• Buckling
• Mixed mode of buckling and crushing
Q.19 What you understand by web crippling?
Solution:
Web crippling is the localized failure of a beam web due to introduction of an excessive
load over a small length of the beam. It occurs at point of application of concentrated
load or at point of support in a beam.
Q.20 What is plasticity index?
Solution:
(b) At support
(a) Below load
Applied load
Reaction
(b)
ENGINEERS PRIDE Class Room/Office Address-C-225, Ganesh Marg, C-Block, Mahesh Nagar
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P a g e 7 | 27
Plasticity Index: It indicates the range of water content over which the soil remains in
plastic state. It is equal to the difference of liquid limit and plastic limit i.e.
Plasticity index ( ) ( )p L pI w w= −
Where;
Lw = Liquid limit of the soil
pw = Plastic limit of the soil
PART-B
Marks: 60
Note- Attempt all the twelve questions. Each question
carries 5 marks. Answer should not exceed 50 words
Q.21 The maximum shear stress in a solid shaft of circular cross-section having diameter `d’
and objected to a torque T is . If the torque is increased by four times and the diameter
of shaft is increased by two times, then find the maximum shear stress in the shaft.
Solution:
We know that
Pl
T
R
=
2
1
P1 1 1
2 2 2 P
l
l
R T
R T
=
4
1 1 1 1
2 1 1 1
2
2 4
R T R
R T R
=
1
2
1 116
2 4
=
ENGINEERS PRIDE Class Room/Office Address-C-225, Ganesh Marg, C-Block, Mahesh Nagar
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and Durga Pura Railway Station), Jaipur, Rajasthan, 9660807149, 7014320833, 8448449932
P a g e 8 | 27
12
2
=
22
=
Q.22 The principal stresses at a point in an elastic material are given as →
1 2 31.5 , , & 0.3, 250 MPayf = = = − = =
For no failure,find the value of as per maximum strain energy theory?
Solution:
As per maximum strain energy theory -
2
2 2 2
1 2 3 1 2 2 3 3 1
12
2 2
yf
E E + + − + +
( ) ( ) ( )2 2 2 2 2 2 21.5 2 1.5 1.5 yf + + − − − −
( )22 2 2 22.25 2 0.3 250 + + − −
2 250 250
4.85
2 113.51 MPa
Q.23 Write between homogeneous, isotropic and anisotropic?
Solution:
Homogeneous material: A material is said to be homogeneous when it has identical
properties at all the points in space or which has a uniform composition throughout the
body.
Isotropic material: A material is said to be isotropic when it has identical properties in all
the directions at a particular point.
Anisotropic material: A material is said to be anisotropic when its one or more properties
get changed with direction at a point within the material.
ENGINEERS PRIDE Class Room/Office Address-C-225, Ganesh Marg, C-Block, Mahesh Nagar
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P a g e 9 | 27
Q.24 The porosity (n) and the degree of saturation (S) of a soil sample are 0.7 and 40%,
respectively. In a 3100 m volume of soil, find the volume (in 3m ) of air is ____.
Solution:
30.7, 40%, 100 , ?aS V m V = = = =
vV
V =
0.7100
vV=
370 mvV =
Now, 1w v a a
v v v
V V V VS
V V V
−= = = −
0.40 1 a
v
V
V= −
30.60 0.60 70 42 maa
v
VV
V= = =
Q.25 Write the assumption Terzaghi's theory for ultimate bearing capacity.
Solution:
Terzaghi's Bearing Capacity Theory: Terzaghi (1943) gave a general theory for the bearing
capacity of soils under a strip footing, making the following assumptions:
• The base of footing is rough.
• The footing is laid at a shallow depth, (fD B ) i.e. shallow foundation.
Air
Solids
Water V
𝑉𝑣 = 𝑉 𝑎 + 𝑉𝑤
ENGINEERS PRIDE Class Room/Office Address-C-225, Ganesh Marg, C-Block, Mahesh Nagar
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P a g e 10 | 27
• The shear strength of the soil above the base of the footing is neglected. The soil
above the base is replaced by a uniform surcharge fD
• The load on the footing is vertical and symmetric. i.e. (moment=0)
• Footing is a strip footing (L>>B).
• The shear strength of the soil is governed by the Mohr-Coulomb equation.
• Failure is general shear failure.
• Ground is horizontal.
Q.26 Write difference between compaction and consolidation
Compaction
• Almost an instantaneous phenomenon.
• Soil is always unsaturated.
• Densification is due to a reduction in the volume of air voids at given water
content.
• Specified compaction techniques are used in this process.
Consolidation
• It is a time dependent phenomenon.
• Soil is completely saturated.
• Volume reduction is due to expulsion of pore water from voids.
• Consolidation occurs on account of a load placed on the soil.
Q.27 What do you mean by Influence Line of a stress-resultant in a structure at a given
point/section? Draw ILD for BM at quarter span in a simply supported beam?
Solution:
Water
Water
Solids Solids
𝑉2 𝑉1
∆𝑉
(a) Before consolidation (b) After consolidation
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An influence line represents the variation of reaction, shear, moments or deflection at a
specified point in a member as unit load moves across the member.
Influence line for BM at quarter span in a simple supported beam
By Muller Breslau Principal release moment at C. Introduce unit reaction at C
1A B + = , 13
4 4
y y
L L+ =
44
3y y L+ = ,
16
3y L=
3
16
Ly =
Q.28 Explain about principle of superposition?
Solution:
Total displacements (or) stresses at a point in a structure subjected to several external
loading can be determined by adding the displacement (or) stresses caused by each of
the external loads acting separately.
Requirements for principal of Superposition are
A B
C
L
L/4
L/4 3L/4
D
𝜃𝐴 𝜃𝐵
L/4
L
ENGINEERS PRIDE Class Room/Office Address-C-225, Ganesh Marg, C-Block, Mahesh Nagar
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1. Linear elastic response i.e., Hooke's law is valid. (I.e. Load ∝ displacement)
2. Small displacement theory applies i.e., geometry of the structure must not undergo
significant change when loads are applied.
Q.29 What is plate girder and gantry girders?
Solution:
Plate girder
When the span of beam increases (more than 15m) so that available rolled steel sections
are not able to provide the required section modulus then one has to go for plate girders.
Plate girder is an arrangement of plates that provides the required section modulus when
the span is large. These are particularly useful for spans ranging from 15 m to 50 m.
Gantry girders
Gantry girders are laterally unsupported beam to carry heavy loads from place to place at
the construction sites, mostly these are of steel material. A girder is a support beam used
in construction. It is the main horizontal support of a structure which supports smaller
beams.
Q.30 Write the principal step in the design of a steel member subjected to axial tension?
Solution:
Steps of designing: -
(1) Permissible stress in tension member = 0.6 yf = at
End carriage
Crab
Gantry girder
Gantry girder
Load
Crane bridge
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P a g e 13 | 27
(2) Calculate net area of tension member = 0.6
net
y
PA
f=
(3) Increase netA by 40% to get and required after deduction for rivet holes.
(4) Choose appropriate section from steel table
(5) Check for netA
(6) netA (Provided) netA (calculated)
(7) Provide connections.
Q.31 Differentiate between Limit state method and Working state method.
Solution:
Working state method Limit state method
1 It is deterministic approach. 1 It is probabilistic approach.
2 In this method safety & serviceability
are checked at working load
2 In this method safety is checked at
ultimate load & serviceability is
checked at working load
3 In WSM, We underestimate the
material strength & don’t
overestimate the load
3 In LSM, We underestimate the
material strength & overestimate
the load
Q.32 What are the assumptions made for the limit state collapse in compression and write an
expression for factored axial load for a short column?
Solution:
Assumption: -
(1) Plane section before bending remains plane after bending
(2) Strain diagram is linear.
(3) Maximum permissible stress in concrete is 0.45fck and in steel 0.67fy.
(4) Maximum strain in concrete in direct compression is 0.002 and in case of bending is
0.0035.
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(5) The maximum compressive strain in highly compressed extreme fibre in concrete
subjected to axial compression and bending but when there is no tension on the section,
is taken as 0.0035 minus 0.75 times the strain at the least compressed extreme fibre.
Expression for factored axial load for a short column:-
0.4 0.67 u ck c y scP f A f A= +
uP = Factored load for short column
ckf = characteristic strength of concrete
yf = characteristic strength of steel
cA = area of concrete
scA = Area of longitudinal reinforcement.
PART-C
Marks: 100
Note- Attempt any 5 out of 7 questions. Each question
carries 20 marks. Answer should not exceed 200 words
Q.33 A beam fixed at one end and simply supported at the other end is having a hinge at B as
shown in the figure. Determine the deflections (a) under the load and (b) at the hinge B.
Use moment area method.
Solution:
A B
C D
40 kN
2 m 2 m 2 m
𝐸𝐼 = constant
Hinge
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P a g e 15 | 27
Taking member BC
By symmetry 20 kNB CR R= =
Thus, FBD of the given beam
Thus, BMD
EI of the given beam
According to moment area method, deflection
at the hinge point B ( )B is
/B A A B AAB = + +
( )1 40 2 160
2 22 3 3EI EI
= − = −
A B
C D
40 kN
2 m 2 m 2 m
𝐸𝐼 = constant
C
𝑅𝐵
D
40 kN
2 m 2 m
𝑅𝐶
20 kN
B A C
D
40 kN
20 kN 20 kN
B
A B C D
40
𝐸𝐼 kN-m 2 m 2 m
2 m 40
𝐸𝐼 kN-m
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Deflection under the load
/ 0BCC B B C BBC = + + =
160 1 40
4 4 2 03 2BCBEI EI
− + + =
160 1604
3BCBEI EI
= −
40 40 80
3 3BCBEI EI EI
= − = −
/BCD B B D BBD = + +
160 80 1 40 2 160 160 802
3 3 2 3 3 3 3EI EI EI EI EI EI
= − − + = − − +
( )320 80 240 80
3 3EI EI EI
− += = − = −
Deflection shape of the beam will be
Q.34 A vertical cylindrical steel storage tank has 30 m diameter and the same is filled up to a
depth of 15 m with the gasoline of relative density 0.74. If the yield stress for steel is 250
MPa, find the thickness required for the wall plate. Adopt a factor of safety of 2.5 and
neglect localized bending effects, if any.
Solution:
2° curve 2° curve
ENGINEERS PRIDE Class Room/Office Address-C-225, Ganesh Marg, C-Block, Mahesh Nagar
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Density of gasoline = 0.74 × 1000 = 740 kg/m3
Hoop stress ( )2
h
pd
t =
where 740 9.81 15p gh= =
5 21.089 10 N/m=
20.1089 N/mm=
( ) 2
0.1089 30000 250 N/mm
2h
t
=
6.534 mmt …(i)
Longitudinal stress ( )( ) 2
0.1089 30000 250 N/mm
4 4t
pd
t t = =
t 3.267 mm …(ii)
From (i) and (ii), t 6. 534 mm
Using a FOS of 2.5, t 6.534 × 2.5 = 16.335 mm
Adopt t = 18 mm
Q.35 A 3.0 m square footing is located in a dense soil and at a depth of 2.0 m. Determine the
ultimate bearing capacity for the following water table positions:
(i) at ground surface,
(ii) at footing level and
(i) at 1m below the footing.
30 m
15 m
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P a g e 18 | 27
The moist unit weight of sand above the water table is 18 kN/m3 and the saturated unit
weight is 3
q20 kN/m , 35°, c 0, N 33 and N 34.0. = = = =
Solution:
Given data:
Width of square footing, B = 3.0m
Depth of footing, 2.0 mfD =
Bulk unit weight of sand, 318 kN/mt =
Saturated unit weight of sand, 320 kN/msat =
q35°, c 0, N 33 and N 34.0 = = = =
We know that for a square footing, ultimate bearing capacity is
1.3 0.4 u c f qq cN D N BN = + +
For the given soil which is sand c = 0
0.4 u f qq D N BN = +
The value of in the above equation is susceptible to the location of water table. Thus
for each water table position, the ultimate bearing capacity will be different. So, the
ultimate bearing capacity will be given as
1 20.4 u f qq D N BN = +
(i) When water table is at the ground surface
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Now for the given condition, soil above the footing and below the footing is submerged,
so
1 220 9.81 and 20 9.81 = − = −
3 3
1 210.19 kN/m and 10.19 kN/m = =
1 20.4u f qq D N BN = +
210.19 2 33 0.4 10.19 3 34 1088.29 kN/m= + =
(ii) When the water table is at the footing level
Now for the given condition, soil above the footing is moist soil and below the footing is
submerged, so
1 2 and sat w = = −
3 3
1 218 kN/m and 20 9.81 10.19 kN/m = = − =
1 20.4u f qq D N BN = +
18 2 33 0.4 10.19 3 34= +
21603.75 kN/m=
(iii) When the water table is at 1 m below the footing
Now for the given condition, soil above the footing is moist and below the footing it is
moist up to 1m and below this, it is submerged, so
𝐷𝑓 𝐷𝑤1
B 𝐷𝑤2
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3
1 218 kN/m and B 1 18 2 (20 9.81) 38.38 = = + − = KN/m2
1 20.4u f qq D N B N = +
218 2 33 0.4 34 1709.968 kN/m= + =
Q.36 A retaining wall 6 m high supports earth with its face vertical. The earth is cohesionless
with particle specific gravity 2.69, angle of internal friction 35° and porosity 40.5%. The
earth surface is horizontal and level with the top of the wall. Determine the earth thrust
and its line of action on the wall if the earth is water logged to level 2.5 m below the top
surface. Neglect wall friction. Draw the pressure diagrams.
Solution:
Given data:
Height of retaining wall, H = 6 m
Specific gravity of soil particles, G = 2.69
Angle of internal friction, 35 =
Porosity, n = 40.5% = 0.405
Depth of water table below top surface = 2.5 m
We know that 1
en
e=
+
0.405
0.6811 1 0.405
ne
n= = =
− −
32.69 0.6819.81 19.67 kN/m
1 1 0.681sat w
G e
e
+ + = = =
+ +
32.69 9.8115.7 kN/m
1 1 0.681
wdry
G
e
= = =
+ +
3' 19.67 9.81 9.86 kN/msat w = − = − =
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P a g e 21 | 27
1 sin 1 sin 350.271
1 sin 1 sin 35aK
− − = = =
+ +
for z = 2.5 m
20.271 15.7 2.5 10.64 kN/mv aP K z= = =
Total active thrust for z = 2.5 m is given by
1
1 12.5 10.64 2.5 13.3 kN/m
2 2vP P= = =
For z = 3.5 m
1
2' 0.271 9.86 3.5 9.35 kN/mv aP K z= = =
2
29.81 3.5 34.335 kN/mv wP z= = =
1 2
29.35 34.335 43.685 kN/mv vP P+ = + =
Total active thrust for z = 3.5 m is given by
( )2 3
143.685 3.5 10.64 3.5
2P P+ = +
76.45 37.24 113.69 kN/m= + =
Total active thrust on the retaining wall =
1 2 3P P P+ + = 13.3 + 76.45 + 37.24 = 126.99 kN/m
Location of total thrust from base =
H = 6 m
2.5 m
3.5 m
10.64 9.35 34.335
𝑃1
𝑃2
𝑃3
𝛾𝑑𝑟𝑦 = 15.7 kN/m3
𝛾′ = 9.86 kN/m3
ENGINEERS PRIDE Class Room/Office Address-C-225, Ganesh Marg, C-Block, Mahesh Nagar
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P a g e 22 | 27
2.5 3.5 3.513.3 3.5 37.24 76.45
3 2 31.67m
126.99
+ + + =
Q.37 A continuous beam ABCD consists of three spans, and is loaded as shown in Figure. Ends
A and D are fixed. Determine the bending moments at the supports and plot the bending
moment diagram.
Solution:
(a) Fixed end moments (kN-m units)
22 66
12FABM
= − = −
22 66
12FBAM
= + = +
2
2
5 3 22.4
5FBCM
= − = −
2
2
5 2 33.6
5FCBM
= + = +
8 55
8FCDM
= − = −
8 55
8FDCM
= + = +
(b) Distribution factors:-
The relative stiffiness and distribution factors are calculated in Table
Joint Member Relative stiffness Sum
Distribution
6 m 5 m 5 m
3 m 2 m 2.5 m 2.5 m
A B C 2 kN/m
5 kN 8 kN
D
I 2I I
ENGINEERS PRIDE Class Room/Office Address-C-225, Ganesh Marg, C-Block, Mahesh Nagar
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P a g e 23 | 27
B BA
BC
𝐼
6
2𝐼
5
17𝐼
30
5
17
12
17
C CB
CD
2𝐼
5
𝐼
5
3𝐼
5
2
3
1
3
(c) Moments distribution:- The moment distribution is carried out in Table. The final
bending moment diagram and the deflected shape of the beam are shown in figure (b)
and (c) respectively.
-6.00 +6.00
-1.06
-2.40 +3.60
-2.54 +0.93
-5.00 +5.00
+0.47
F.E.M
Balance
-0.53
-0.14
0.46 -1.27
-0.32 +0.85
+0.23
+0.42
Carry over
Balance
-0.07
-0.12
+0.42 -0.16
-0.30 +0.11
+0.21
+0.05
C
B
-0.06
-0.014
+0.05 -0.15
-0.03 +0.10
+0.03
+0.05
C
Balance
-6.66 +4.66 -4.66 +4.01 -4.01 +5.47 Final moments
A B
C D
5
17
12
17
2
3
1
3
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P a g e 24 | 27
Q.38 Mention the various types of losses in prestressing members in pretensioning and post
tensioning. Explain them very briefly
Solution:
The losses in prestressed members are:
pre-tensioned concrete post tensioned concrete
1. Elastic deformation of concrete 1. No loss due to elastic deformation if all
the tendons are tensioned simultaneously.
2. Shrinkage of concrete 2. Shrinkage of concrete
3. creep of concrete 3. creep of concrete
4. Relaxation of steel 4. Relaxation of steel
5.Frictional and wobbling losses
6. Anchorage slip
1. Loss due to elastic deformation of concrete: This loss depends on the modular ratio and
average stress in concrete at the level of steel. In case of post tensioned beams if all the
wires are simultaneously tensioned no loss due to elastic deformation occurs. But if wires
are successively tensioned, there will be loss of prestress due to elastic deformation.
Loss = m.fc
6.66
6 m
4.01kN-m 5.47
5 m 5 m
3 m 2 m 2.5 m 2.5 m A
B (a) C 2𝐼 2 kN/m
5 kN 4.66kN-m 8 kN
D 𝐼
6 10 9
6.66 kN-m
(b)
A B
(c) C D
4.66kN-m 5.47kN-m 4.01kN-m
- + -
+
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P a g e 25 | 27
2. Loss due to shrinkage of concrete: This shrinkage of concrete in prestressed members
results in shortening of tensioned wires and hence contributes to the loss of stress.
Loss = Shrinkage strain × sE
3. Loss due to creep of concrete: The sustained prestress in the concrete of a prestressed
member results in creep of concrete which effectively reduces the stress in high tensile
steel.
Loss . . cm f=
4. Loss due to relaxation of stress in steel: It is generally equal to 1 to 5% of prestress.
5. Loss of stress due to friction: In case of post tensioned members, the tendons are housed
in ducts preformed in concrete. The ducts are either straight or follow curved profile
depending upon the design requirements. Consequently on tensioning the curved
tendons, loss of stress occurs in post-tensioned members due to friction between the
tendons and the surrounding concrete ducts.
Prestressing force at a distance x from jacking end ( )
0
kx
xP Pe − +=
6. Loss due to anchorage slip: In most post-tensioning systems, when the cable is tensioned
and the jack is released to transfer prestress to concrete, the friction wedges, employed
to grip the wires, slip over a small distance before the wires are firmly housed between
the wedges. The magnitude of slip depends upon the type of wedge and the stress in
wires
Loss = s
LE
L
Q.39 Two wheels, placed at a distance of 2.5 m part, with a load of 200 kN on each of them,
are moving on a simply supported girder (𝐼-section) of span 6.0 m from left to right. The
top and bottom flanges of the 𝐼-section are 200 × 20 mm and the size of web plate 800
× 6 mm. If the allowable bending compressive, bending tensile and average shear
stresses are 110 MPa, 165 MPa and 100 MPa respectively, check the adequacy of the
section against bending and shear stress. Self weight of the girder may be neglected.
Solution:
Maximum S.F. will occur when left 200 kN load is at A or right 200 KN load is at B.
ENGINEERS PRIDE Class Room/Office Address-C-225, Ganesh Marg, C-Block, Mahesh Nagar
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P a g e 26 | 27
( )6 200 6 200 6 2.5AR = + −
AR = 316.67 kN
Maximum S.F = 316.67 kN
Maximum B.M. will occur when one of 200 kN load and resultant of both 200 kN loads
are equidistant from center of span as shown.
Max. BM under left 200 kN load = AR (3−0.625)
= 158.33 (2.375) = 376.034 kNm
A B
6 m
1
1
+
−
1.25 m 1.25 m
2.5 m 200 kN 200 kN
ILD of S.F. at A
ILD of S.F. at B
ILD for Maximum BM
A B
0.79167
0.625 m 1.875 m
6 m
ENGINEERS PRIDE Class Room/Office Address-C-225, Ganesh Marg, C-Block, Mahesh Nagar
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P a g e 27 | 27
3 3840 800200 194
12 12xxI = −
6 41601.067 10 mm=
Maximum bending compressive stress 6
6
376.034 10420
1601.067 10
=
2 2 98.64 N/mm < 110 N/mm= (OK)
Maximum bending tensile stress = 6
6
376.034 10400
1601.067 10
2 2 98.64 N/mm < 165 N/mm= (OK)
Maximum shear stress ( )2 2 2
8 w
F BD d d
I t
= − +
( )3
2 2 2
6
316.67 10 200840 800 800
8 1601.067 10 6
= − +
2 2 69.88 N/mm < 100 N/mm= (OK)
Section is safe against bending and shear
200
200
800
6
20
20
X X