Question Paper Nov-Dec-2018.pdf

Question Paper Code: 20282
B.E./B.Tech. DEGREE EXAMINATION, NOVEMBER/DECEMBER 2018.
Seventh Semester
Civil Engineering
CE 6701 — STRUCTURAL DYNAMICS AND EARTH QUAKE ENGINEERING
(Regulations 2013)
(Common to PTCE 6701 — Structural Dynamics and Earth Quake Engineering for
B.E (Part-Time) — Fifth Semester — Civil Engineering - Regulations 2014)
Time: Three hours Maximum: 100 marks
Answer for ALL questions.
PART A — (10 x 2 = 20 marks)
1. Explain De Alembert’s principle
According to Newton‘s law F = ma
 D‘Alembert‘s principle which state that a system may be in dynamic
equilibrium by adding to the external forces, an imaginary force, which is
commonly known as the inertia force.
F – ma = 0
F Real Force
ma Imaginary Force
2. Brief about the SDOF system and free body diagram.
 At any instant of time, the motion of this system can be denoted by single
coordinate (x in this case).
 It is represented by a rigid mass, resting on a spring of stiffness ‗k‘ and coupled
through a viscous dashpot (representing damping) having constant ‗c‘. Here, the
mass ‗m‘ represents the inertial effects of damping (or energy dissipation) in the
system.
 Using the dynamic equilibrium relation with the inertial force included,
according to D‘Alembert‘s principle, it can be written as
FI + FD + FS = P(t)
(Inertial Force) (Damping Force) (Elastic Force) (Application Force)
This gives
 ̈ ̇ x, ̇, ̈ respectively denote the displacement, velocity and
acceleration of the system. P (t) is the time dependent force acting on the mass.
9 2 3 1 1 5 1 0 3 x x x
Reg. No

 The above equation represents the equation of motion of the single degree freedom
system subjected to forced vibrations.
3. What is meant by mode super position technique?
Mode super position technique is the method of using the natural frequencies and modes
shapes from the modal analysis to characterize the dynamic response of a structure to
transient or steady harmonic excitations.
4. Enumerate properties of stiffness matrix and mass matrix.
Stiffness Matrix
The matrix [ ] is called stiffness matrix and it also denoted by [k]
Mass Matrix
The matrix * + is called mass matrix and it can also be represented as [m]
 Mij a typical element of matrix M is defined as the force corresponding to co -
ordinate i due as the force corresponding to coordinate i due to unit acceleration
applied to the co-ordinate j.
[M]{Y}+[C]{Y}+[K]{Y} = {P(t)}
5. Classify the faults based on the direction of the movement of blocks.
 Dip-Slip fault
 Strike-Slip fault
 Oblique-Slip fault
6. What is meant by reservoir induced Earthquakes?
 The phenomenon of dam-triggered earthquakes is known as reservoir-induced
seismicity.
 When a dam is built and the reservoir filled with water, the amount of pressure
exerted on the earth in that area changes dramatically.
7. Brief short column damage in RC buildings.
 During past earthquakes, reinforced concrete frame buildings that have columns of
different heights within one storey, suffered more damage in the shorter columns as
compared to taller columns in the same storey.
 This behaviour is called Short Column Effect.
(Or)
 Short column effect arises when a column in a RC frame building is restricted from
moving owing to any obstruction. The obstruction can be:
 Presence of unreinforced masonry infills of partial height of adjoining RC column

 Conditions arising from sloping ground, when some basement columns are shorter
than others,
 Presence of a mezzanine slab (which meets the columns at an intermediate height
between the usual beam-slab system of the floors in RC buildings)
 Presence of a staircase beam/slab or K-braces on building columns (which meets the
columns at an intermediate height between the usual beam-slab system of the floors in
RC buildings
 Presence of a plinth beam making the height of the column below it to be shorter than
that of the column above.
8. Brief P- delta effects
 When a model is loaded, it deflects. The deflections in the members of the model may
induce secondary moments due to the fact that the ends of the member may no longer
be vertical in the deflected position.
 These secondary effects for members can be
accurately approximated through the use of
P-Delta analysis.
 This type of analysis is called ―P-Delta‖
because the magnitude of the secondary
moment is equal to ―P‖, the axial force in the
member, times ―Delta‖, the distance one end of the member is offset from the other
end.
9. Define structural plan density
 The area of vertical members of a building has been reduced drastically from about
50-60% of the plinth area in historic masonry buildings to a meagre 2-4% in modern
RC frame buildings.
 This sharp reduction was possible by the advent of materials whose strength and
stiffness properties are at least one order of magnitude higher.
 This ratio of the area of footprint of vertical elements resisting the lateral load and the
plinth area of the building is called Structural Plan Density (SPD) of the building.
10. Differentiate Weak storey and soft storey.
According to UBC code
Soft storey = the storey which the lateral stiffness is less than 70% of the stiffness
of the storey above.
Weak storey = the storey which the story strength is less than 80% of that the
storey above.

PART B - (5 x 13 = 65 marks)
11. (a) A machine of mass one tonne is acted upon by an external force of 2450 N at a
frequency of 1500 rpm. To reduce the effects of vibration, isolator of rubber having a
static deflection of 2 mm under the machine load and an estimated damping =0.2 are
used. Determine
(i) The force transmitted to the foundation (6)
(ii) The amplitude of vibration of Machine (7)
To find:
i. Force transmitted to the foundation
ii. Amplitude
Given Data:
i. Mass of the machine = 1tonne = 1000kg
ii. External force(Fo) = 2450N
iii. Frequency = 1500rpm
iv. Static deflection(Xst) = 2mm = 2x10-3
m
v. Damping(ρ) = 0.2
Solution:
i.The force transmitted to the foundation
To find necessary data,
a. Frequency
;
rad / sec
b. Spring Stiffness,
;
k = 4.905 x 106
N/m
c. Natural Frequency
√ ; √
= 70.03 rad/sec
d. Frequency ratio
;
e. The Transmissibility is,

√
√
√
√
The transmitted force to the foundation,
F1 = Tr x Fo
F1 = 0.078 x 2450
F1 = 191.1 N
ii.The amplitude of vibration of Machine
⁄
√
⁄
√
= X = 2.936 x 10-5
X = 0.02936 mm
(Or)
(b) A SDOF system having a mass of 2.5 kg is set into motion with viscous damping and
allowed to oscillate freely. The frequency of oscillation is found to be 20 Hz and the
measurement of amplitude of vibration shows two successive amplitudes to be 6 mm and 5.5
mm. Determine the viscous damping coefficient. (13)
To Find:
i. Viscous damping coefficient
Given Data:
i. Mass of the system(m) = 2.5kg
ii. Frequency(fd) = 20 Hz
iii. Two Successive amplitude
X1 = 6mm
X2 = 5.5mm
Solution:
i.To Find necessary data,
a. Logarithmic decrement,

= 0.087
b. Damping ratio,
;
ρ = 0.014
c. Damped natural frequency,
;
d. Natural frequency
√ ;
√
√
e. Spring force
For small value of,
√ ; ;
ii.To Find Viscous damping coefficient
;
√
; √
√
C = 6.41 N-s/m

12. (a) Determine the natural frequencies of vibration of MDOF system by using
matrix method as shown in figure 12 (a). (13)
figure 12 (a)
Solution:
i). To draw spring mass system:
ii). To draw free body diagram:
iii). To write equation of motion
For mass 1,
̈ =0
̈ =0
̈ =0…… (1)
For mass 2,
̈ + - =0
̈ =0
̈ - =0…… (2)
For mass 3,
̈ + =0
̈ + =0…... (3)
iv). To form Matrix equation:
[ ] {
̈
̈
̈
} [ ] { } =0…... (4)
v). To form Characteristic equation:
| | .….. (5)

|[ ] [ ] | ….... (6)
In our problem says m1 = m2=m3=m & k1=k2=k3=k
|[ ] [ ] |
|[ ] [ ]|
[ ] …... (7)
Expanding above equation we,
{ [ ]} { }
{ [( ) ]} { }
{ } { }
….... (8)
Multiplying by –ve sign we get,
….... (9)
We take & above equation rewritten as,
….... (10)
Solved the above characteristic equation we get,
λ1= 3.247k λ2=0.198k λ3 = 1.555k
WKT,
λ1 = 3.247k λ2 =0.198k λ3 = 1.555k
= = =
= 1.802√ rad/s = 0.445√ rad/s = 1.247√ rad/s
vi). To find Mode Shapes
By using static condensation method,
[ ] { } = { } ….... (11)

, - [ ] , - = , -
, - [ ] = , - ….... (12)
For first mode λ1 = 3.247k,
[ ] , - = { }
* + , - ={ }
* + , - ={ }
, - ={ }….... (13)
Eigen vectors for first mode of vibrations is,
{ } ={ }
For second mode λ2 =0.198k
[ ] , - ={ }
[ ] , - ={ }
* + , - ={ }
, - = { } ….. (14)
Eigen vectors for second mode of vibrations is,
{ } ={ }
For third mode λ3 = 1.555k,
[ ] , - = { }
[ ] , - = { }
* + , - = { }

, - = { } ….. (15)
Eigen vectors for third mode of vibrations is,
{ } ={ }
vii). To draw mode shape diagram,
viii).Result:
a). Eigen Values,
λ1 = 3.247k λ2 =0.198k λ3 = 1.555k
b) Natural Frequencies,
= 1.802√ rad/s = 0.445√ rad/s = 1.247√ rad/s
c). Eigen Vectors,
{ } ={ }; { }={ } ; { }= { }
(Or)
(b) Determine the natural frequencies of the system as shown in figure 12 (b) by Holzer
method (13)
figure 12 (b)
Solution:
(i) To find spring forces:

Spring in parallel,
⁄ ⁄
(ii) To draw spring mass system:
(iii)To draw free body diagram:
(iv)To write equation of motion
Equation of motion for this system is,
For Mass 1,
̈ = 0
̈ = 0
̈ = 0..... (1)
For Mass 2,
̈ + = 0
̈ + = 0..... (2)
For Mass 3,
̈ + = 0
̈ = 0..... (3)
(v) To form Matrix equation:
[ ] {
̈
̈
̈
} [ ] { } = 0 …... (4)
(vi)To form Characteristic equation:
| | ….. (5)
|[ ] [ ] | ….. (6)
Assume & substitute,
; ; ; ;

|[ ] [ ] |
|[ ] [ ] |
[ ] …. (7)
{
{ [ ]}
{ [ ]}
{ [ ]}
}
{
{ [ ]}
{ [ ]}
{ [ ]}
}
{
[ ]
[ ]
[ ] }
{
[ ] [ ]
[ ]
[ ] }
….(8)
Multiplying by –ve sign we get,
....(9)
We take & above equation rewritten as,
... (10)
Solved the above characteristic equation we get,
λ1= 5.730k λ2=0.614k λ3 = 2.073k
(vii) To find natural frequency
WKT,
λ1 = 5.730k λ2 =0.614k λ3 = 2.073k
= = =
= 2.394√ rad/s = 0.784√ rad/s = 1.440√ rad/s

13. (a) (i) Explain elastic rebound theory. (5)
The concept of possible mode of origin of tectonic earthquakes is known as
elastic rebound theory.
According to professor , of Geology H.F.Reid, materials of the earth being
elastic, can withstand a certain amount of stress without deforming permanently, but
if the stress is continued for a long period of time, or if it is increased in magnitude,
the rocks will first take a permanent deformation or strain and eventually rupture.
A fault is a break or fracture in the material of the earth along which there has
been displacement. When the rupture occurs, rocks on either side of the fault tent to
return to their original shape because of their elasticity and an elastic rebound occurs.
This rebound sets up the seismic waves. The earthquake mechanism may be
explained as occurring in three phases in accordance with elastic rebound theory.
1. The preparing process
2. The rupture phase
3. The post failure adjustment
The elastic rebound theory is an explanation for how energy is spread during
earthquakes. As rocks on opposite sides of a fault are subjected to force and shift, they
accumulate energy and slowly deform until their internal strength is exceeded. At that
time, a sudden movement occurs along the fault, releasing the accumulated energy,
and the rocks snap back to their original un-deformed shape.
In geology, the elastic rebound theory was the first theory to satisfactorily
explain earthquakes. Previously it was thought that ruptures of the surface were the
result of strong ground shaking rather than the converse suggested by this theory.
The theory explained

Following the great 1906 San Francisco earthquake, Harry Fielding Reid
examined the displacement of the ground surface around the San Andreas Fault.
From his observations he concluded that the earthquake must have been the
result of the elastic rebound of previously stored elastic strain energy in the rocks on
either side of the fault.
In an inter-seismic period, the Earth's plates (see plate tectonics) move relative
to each other except at most plate boundaries where they are locked.
Thus, if a road is built across the fault as in the figure panel Time 1, it is
perpendicular to the fault trace at the point E, where the fault is locked.
The far field plate motions (large arrows) cause the rocks in the region of the
locked fault to accrue elastic deformation, centimetres per year, over a time period of
many years. When the accumulated strain is great enough to overcome the strength of
the rocks, an earthquake occurs.
During the earthquake, the portions of tire rock around the fault that were
locked and had not moved 'spring' back, relieving the displacement in a few seconds
that the plates moved over the entire inter seismic period (D1 and D2 in Time 3). The
time period between Time 1 and Time 2 could be months to hundreds of years, while
the change from Time 2 to Time 3 is seconds.
Like an elastic band, the more the rocks are strained the more elastic energy is
stored and the greater potential for an event. The stored energy is released during the
rupture partly as heat, partly in damaging the rock, and partly as elastic waves.
Modem measurements using GPS largely support Reid‘s theory as the basis of
seismic movement, though actual events are often more complicated.
(ii) Explain the seismic susceptibility of Indian Subcontinent (8)
RECENT SEISMIC ZONES IN INDIA: (BIS-2002)
 The 1993 Latur earthquake of magnitude 6.3 caused intensity IX damages
but prior to the earthquake, Latur was placed in seismic zone 1 where no
such magnitude of earthquake was expected.
 The Latur earthquake further led to the revision of the seismic zonation
map of India. The map was revised again in 2002 with only four zones
such as II, III, IV and V (IS: 1893 (Part 1): 2002).
 The Peninsular India was modified and Zones I and II were combined. The
new zone placed the 1993 Latur earthquake in zone III.
 The areas falling under zone V is most seismically active. The areas under
this zone are the entire northeastern part of India, parts of northwestern
Bihar, the Kangra Valley in Himachal Pradesh, Andaman and Nicobar
Islands, eastern part of Uttaranchal, the Rann of Kutchh in Gujarat and the
Srinagar area in Jammu and Kashmir.

 Two major metropolitan cities, with a high population density, i.e. Delhi,
lie in zone IV, and Kolkata, at the boundary of zone III and IV of the
zonation map.
 The recent four seismic zones of India are assigned PGA values ranging
from 0.1 g to 0.4 g (Giardini et al. 1999) with 10% probability of
exceedance in 50 years.
 The changes in the zonation map of India with the occurrence of
significant earthquakes are an indication that the zoning at a national level
does not provide the solution for tackling the seismic hazards.
 Zone 5
o Zone 5 covers the areas with the highest risks zone that suffers
earthquakes of intensity MSK IX or greater.
o The IS code assigns zone factor of 0.36 for Zone 5.
o Structural designers use this factor for earthquake resistant design
of structures in Zone 5.
o The zone factor of 0.36 is indicative of effective (zero period) level
earthquake in this zone. It is referred to as the Very High Damage
Risk Zone.
o The region of Kashmir, the Western and Central Himalayas, North
and Middle Bihar, the North-East Indian region, the Rann of
Kutch and the Andaman and Nicobar group of islands fall in this
zone.
o Generally, the areas having trap rock or basaltic rock are prone to
earthquakes.
 Zone 4
o This zone is called the High Damage Risk Zone and covers areas
liable to MSK VIII.
o The IS code assigns zone factor of 0.24 for Zone 4 Jammu and
Kashmir, Himachal Pradesh, Uttarakhand, Sikkim, the parts
of Indo-Gangetic plains (North Punjab, Chandigarh, Western Uttar
Pradesh, Terai, North Bengal, Sundarbans) and the capital of the
country Delhi fall in Zone 4.
o In Maharashtra, the Patan area (Koynanagar) is also in zone no-4.
In Bihar the northern part of the state like Raxaul, Near the border
of India and Nepal, is also in zone no-4.
 Zone 3
o This zone is classified as Moderate Damage Risk Zone which is
liable to MSK VII. and also 7.8
o The IS code assigns zone factor of 0.16 for Zone 3.
 Zone 2
o This region is liable to MSK VI or less and is classified as the Low
Damage Risk Zone.
o The IS code assigns zone factor of 0.10 (maximum horizontal
acceleration that can be experienced by a structure in this zone is
10% of gravitational acceleration) for Zone 2.

 Zone 1
o Since the current division of India into earthquake hazard zones
does not use Zone 1, no area of India is classed as Zone 1.
o Future changes in the classification system may or may not return
this zone to use.
(Or)
(b)Discuss ground subsidence, slope instability due to earthquake and method of
evaluating liquefaction potential. (13)
Ground Subsidence
Land subsidence can occur in various ways during an earthquake. Movement
that occurs along faults can be horizontal or vertical or have a component of both. As
a result, a large area of land can subside drastically during an earthquake. Land
subsidence can also be caused during liquefaction.
Groundwater-related subsidence is the subsidence (or the sinking) of land resulting
from groundwater extraction. It is a growing problem in the developing world as cities
increase in population and water use, without adequate pumping regulation and
enforcement.
The probable cause was declining groundwater levels. Here are some other
things that can cause land subsidence: The principal causes are aquifer-system
compaction, drainage of organic soils, underground mining, hydro compaction,
natural compaction, sinkholes, and thawing permafrost.
 Subsidence caused by aquifer system compaction due to the lowering of
ground-water levels by sustained ground-water overdraft
 Subsidence caused by the hydro compaction of moisture-deficient deposits
above the water table
 Subsidence from oil and gas field withdrawal
 Subsidence related to crustal geo tectonic movements
 Problems caused by subsidence
 Damage to bridges, canals, roads, storm drains, sewers, canals and levees
 Damage to buildings
 Failure in well casings
 In low lying areas subsidence can result in tides moving into areas that
were once above sea level
 Large economic costs

Slope instability
Slope stability is the potential of soil covered slopes to withstand and undergo
movement. Stability is determined by the balance of shear stress and shear strength. Mass
movements can be caused by increase in shear stress, such as loading, lateral pressure, and
transient forces.
14. (a) Elaborate the Planning and Architectural consideration in RC buildings and
discuss the potential deficiencies of buildings exist in our society. (13)
Planning and Architectural consideration in RC buildings as per IS code book
IS1893:1984, IS13920:1993, IS4326:1993 are,
 Lightness
o Since the earthquake force is a function of mass, the building should be as
light as possible consistent with structural safety and functional requirements.
Roofs and upper storeys of buildings in particular should be designed as light
as possible.
 Continuity of Construction
o As far as possible, all parts of the building should be tied together in such a
manner that the building acts as one unit.
o For integral action of building, roof and floor slabs should be continuous
throughout as far as possible.
o Additions and alterations to the structures should be accompanied by the
provision of positive measures to establish continuity between the existing and
FAULTING
Seismic
Energy
Topographic
Relief
Soil Deposits
Surface Rupture Tsunamis
Landslides Liquefaction
Strong Ground
Motion
Amplified ground
shaking
Amplified ground
shaking

the new construction.
 Projecting and Suspended Parts
o Projecting parts should be avoided as far as possible. If the projecting parts
cannot be avoided, they should be properly reinforced and firmly tied to the
main structure.
o Ceiling plaster should preferably be avoided. When it is unavoidable, the
plaster should be as thin as possible.
o Suspended ceiling should be avoided as far as possible. Where provided, they
should be light and adequately framed and secured.
 Shape of Building
o In order to minimize torsion and stress concentration, the building should have
a simple rectangular plan.
o It should be symmetrical both with respect to mass and rigidity so that the
centre of mass and rigidity of the building coincide with each other.
o It will be desirable to use separate blocks of rectangular shape particularly in
seismic zones V and IV.
 Preferred Building Layouts
o Buildings having plans with shapes like, L, T, E and Y shall preferably be
separated into rectangular parts by providing separation sections at appropriate
places typical examples are shown in Fig. 5.1.
Strength in Various Directions
o The structure shall have adequate strength against earthquake effects along
both the horizontal axes considering the reversible nature of earthquake forces.
 Foundations
o For the design of foundations, the provisions of IS 1904: 1986 in conjunctions
with IS 1893: 1984 shall generally be followed.
o The sub-grade below the entire area of the building shall preferably be of the
same type of the soil. Wherever this is not possible, a suitably located
separation or crumple section shall be provided.
o Loose fine sand, soft silt and expansive clays should be avoided. If

unavoidable, the building shall rest either on a rigid raft foundation or on piles
taken to a firm stratum. However, for light constructions the following
measures may be taken to improve the soil on which the foundation of the
building may rest:
a).Sand piling, b) Soil stabilization.
o Structure shall not be founded on loose soil, which will subside or liquefy
during an earthquake resulting in large differential settlement.
 Flat roof or floor
o Flat roof or floor shall not preferably be made of terrace of ordinary bricks
supported on steel, timber or reinforced concrete joists, nor they shall be of a
type which in the event of an earthquake is likely to be loosened and parts of
all of which may fall. If this type of construction cannot be avoided, the joists
should be blocked at ends and bridged at intervals such that their spacing is
not altered during an earthquake.
 Pitched Roofs
o For pitched roofs, corrugated iron or asbestos sheets should be used in
preference to country, Allahabad or Mangalore tiles or other loose roofing
units.
o All roofing materials shall be properly tied to the supporting members.
o Heavy roofing materials should generally be avoided.
 Pent Roofs
o All roof trusses should be supported on and fixed to timber band reinforced
concrete band or reinforced brick band. The holding down bolts should have
adequate length as required for earthquake and wind forces.
o Where a trussed roof adjoins a masonry gable, the ends of the purlins should
be carried on and secured to a plate or bearer which should be adequately
bolted to timber reinforced concrete or reinforced brick band at the top of
gable end masonry.
o At tie level, all the trusses and the gable end should be provided with diagonal
braces in plan so as to transmit the lateral shear due to earthquake force to the
gable walls acting as shear walls at the ends.
o NOTE Hipped roof in general have shown better structural behaviour during
earthquakes than gable ended roofs.
 Jack Arches
o Jack arched roofs or floors where used should be provided with mild steel ties

in all spans along with diagonal braces in plan to ensure diaphragm actions.
 Staircases
o The interconnection of the stairs with the adjacent floors should be
appropriately treated by providing sliding joints at the stairs to eliminate their
bracing effect on the floors.
o Ladders may be made fixed at one end and freely resting at the other.
o Large stair halls shall preferably be separated from rest of the building by
means of separation or crumple section. Three types of stair construction may
be adopted as described below:
 Separated Staircases
o One end of the staircase rests on a wall and the other end is carried by columns
and beams which have no connection
with the floors. The opening at the
vertical joints between the floor and the
staircase may be covered either with a
tread plate attached to one side of the
joint and sliding on the other side, or
covered with some appropriate material
which could crumple or fracture during
an earthquake without causing
structural damage.
o The supporting members, columns or
walls, are isolated from the surrounding floors by means of separation or
crumple sections. A typical example is shown in Fig. 5.2.
 Built-in Staircase
o When stairs are built monolithically with
floors, they can be protected against
damage by providing rigid walls at the
stair opening. An arrangement, in which
the staircase is enclosed by two walls, is
given in Fig. 5. In such cases, the joints,
as mentioned in respect of separated
staircases, will not be necessary.
o The two walls mentioned above,
enclosing the staircase, shall extend

through the entire height of the stairs and to the building foundations.
 Staircases with Sliding Joints
o In case it is not possible to provide rigid walls
around stair openings for built-in staircase or to
adopt the separated staircases, the staircases
shall have sliding joints so that they will not act
as diagonal bracing. (Fig. 4)
 Box Type Construction
o This type of construction consists of prefabricated or in-situ masonry wall along
with both the axes of the building. The walls support vertical loads and also act
as shear walls for horizontal loads acting in any direction.
o All traditional masonry construction falls under this category. In prefabricated
wall construction, attention should be paid to the connections between wall
panels so that transfer of shear between them is ensured.
 Fire Safety
o Fire frequently follows an earthquake and therefore buildings should be
constructed to make them fire resistant in accordance with the provisions of
relevant Indian Standards for fire safety.
o The relevant Indian Standards are IS 1641 : 1988, IS 1642 : 1989, IS 1643 :
1988, IS 1644 : 1988 and IS 1646 : 1986.
(Or)
(b)Discuss the dynamic analysis procedure of RC framed Structure as per IS 1893:2002
with suitable assumed data of your choice. (13)
Assumed data:
Plan Dimensions = 7m
Storey Height = 3.5m
Total weight of Beams in a storey (Wb) = 130 kN
Total weight of Slab in a storey (Ws) = 250 kN
Total weight of Column in a storey (Wc) = 50 kN
Total weight of Walls in a storey (Ww) = 530 kN
Live load (WL) = 130 kN
Weight of terrace floor (Wt) = 655 kN

Solution:
Si. No. Description Result Ref
1 To find Total seismic weight
Wi = Wb + Ws+ Wc + Ww + WL
W1= 130 + 250+ 50 + 530 + 130
W2= 130 + 250+ 50 + 530 + 130
Wt = weight on terrace
W = W1 + W2 + Wt
1090 kN
1090 kN
655 kN
W = 2835 kN
2 To find Natural Period
Assume with brick infill,
√
√ T = 0.357 s
IS1893(I):2002
Pg. No : 24
Clause : 7.6.2
3 To find Design Horizontal seismic
coefficient
a) Z = Zone factor
Bhuj situated in zone V
b) I = Importance factor
Building used as School
building
c) R = Response reduction factor
Special Moment Resisting
Frame
d) =
T = 0.357 s,
Damping = 5%,
Type of rock = Hard rock
Z = 0.36
I = 1.5
R = 5
= 2.5
IS1893(I):2002
Pg. No : 14
Clause : 6.4
Pg. No : 16
Table : 2
&
Annex-E,Pg.No:35
Pg. No : 18
Table : 6
Pg. No : 23
Clause : 7
Pg. No : 16
Clause : 6.4.5
&
Table : 3 for
damping is given
4 To find Base shear
VB = Ah W
VB = 0.135 x 2835 VB = 382.725 kN
IS 1893(I):2002
Pg. No : 24
Clause : 7.5.3
5 To find Equivalent lateral load (or)
Equivalent seismic force
{
∑
}
IS 1893(I): 2002
Pg. No : 24
Clause : 7.7.1

Floor Weight
(Wi)
Height
(hi)
(hi)2
∑
{
∑
}
G.F 1090 3.5 12.25 13,352.5 0.181 69.273 kN
F.F 1090 7 49 53,410.0 0.725 277.476 kN
S.F 655 10.5 110.25 06,877.5 0.093 35.593 kN
73,640.0 0.999 382.342 kN
15. (a)Elaborate the seismic detailing requirements of a shear wall and elements of RC
framed Structures as per IS: 13920-1993. (13)
 Construction of Buildings with Shear Walls preferably in Seismic Regions,
Reinforced concrete (RC) buildings often have vertical plate-like RC walls called
Shear Walls in addition to slabs, beams and columns.
 These walls generally start at foundation level and are continuous throughout the
building height. Their thickness can be as low as
150mm, or as high as 400mm in high rise buildings.
 Shear walls are usually provided along both length
and width of buildings. Shear walls are like
vertically oriented wide beams that carry
earthquake loads downwards to the foundation.
(Fig. 9.32)
 Properly designed and detailed buildings with shear
walls have shown very good performance in past
earthquakes.
 Shear walls in high seismic regions require special detailing.
 Shear walls are efficient, both in terms of construction cost and effectiveness in
minimizing earthquake damage in structural and non-structural elements (like glass
windows and building contents).

o Shear walls provide large strength and stiffness to buildings in the direction of
their orientation, which significantly reduces lateral sway of the building and
thereby reduces damage to structure and its contents.
o Since shear walls carry large horizontal earthquake forces, the overturning
effects on them are large. Thus, design of their foundations requires special
attention.
o Shear walls should be provided along preferably both length and width.
However, if they are provided along only one direction, a proper grid of beams
and columns in the vertical plane
(called a moment-resistant frame)
must be provided along the other
direction to resist strong
earthquake effects.
o Door or window openings can be
provided in shear walls, but their
size must be small to ensure least
interruption to force flow through
walls.
o Shear walls in buildings must be
symmetrically located in plan to reduce ill-effects of twist in buildings. (Fig.
9.33)
o Shear walls are more effective when located along exterior perimeter of the
building – such a layout increases resistance of the building to twisting.
 Ductile Design and Geometry of Shear Walls
o Shear walls are oblong in cross-section, i.e., one dimension of the cross-
section is much larger than the other. While rectangular cross-section is
common, L- and U-shaped sections are also used.
o Overall geometric proportions of the wall, types and amount of reinforcement,
and connection with remaining elements in the building help in improving the
ductility of walls.
o The Indian Standard Ductile Detailing Code for RC members (IS:13920-1993)
provides special design guidelines for ductile detailing of shear walls.
(Or)

(b) Discuss the evaluation of stresses involved in masonry pier with a masonry
building of your choice. (13)
 Protection of Openings in Walls
o Horizontal bands including plinth band, lintel band and roof band are provided
in masonry buildings to improve their earthquake performance.
o Even if horizontal bands are provided, masonry buildings are weakened by the
openings in their walls. Embedding vertical reinforcement bars in the edges of
the wall piers and anchoring them in the foundation at the bottom and in the
roof band at the top, forces the slender masonry piers to undergo bending
instead of rocking.
o In wider wall piers, the vertical bars enhance their capability to resist
horizontal earthquake forces and delay the X-cracking. Adequate cross-
sectional area of these vertical bars prevents the bar from yielding in tension.
Further, the vertical bars also help protect the wall from sliding as well as from
collapsing in the weak direction.
o However, the most common damage, observed after an earthquake, is diagonal
X-cracking of wall piers, and also inclined cracks at the corners of door and
window openings. When a wall with an opening deforms during earthquake
shaking, the shape of the opening distorts and becomes more like a rhombus -
two opposite corners move away and the other two come closer.
o Under this type of deformation, the corners that come closer develop cracks.
The cracks are bigger when the opening sizes are larger. Steel bars provided in
the wall masonry all around the openings restrict these cracks at the corners.
o In summary, lintel and sill bands above and below openings, and vertical
reinforcement adjacent to vertical edges, provide protection against this type of
damage (Fig. 10.13).
 General Principles for Construction of Earthquake Resistant Brick Masonry Building

o Low Strength Masonry constructions should not be permitted for important
buildings.
o It will be useful to provide damp-proof course at plinth level to stop the rise of
pore water into the superstructure.
o Precautions should be taken to keep the rain water away from soaking into the
wall so that the mortar is not softened due to wetness. An effective way is to
take out roof projections beyond the walls by about 500 mm.
o Use of a water-proof plaster on outside face of walls will enhance the life of
the building and maintain its strength at the time of earthquake as well.
o Ignoring tensile strength, free standing walls should be checked against
overturning under the action of design seismic coefficient, ah, allowing for a
factor of safety of 1.5.
 Categories of Buildings
o For the purpose of specifying the earthquake
resistant features in masonry and wooden
buildings, the buildings have been
categorized in five categories A to E based
on the seismic zone and the importance of
building I,
o Where, I = importance factor applicable to
the building [Ref. Clause 6.4.2 and Table - 6 of IS 1893 (Part 1): 2002].
The building categories are given in Table – 10.2.
 Brickwork in Weak Mortars
o The fired bricks should have a compressive strength not less than 3.5 MPa.
Strength of bricks and wall thickness should he selected for the total building
height.
o The mortar should be lime-sand (1:3) or clay mud of good quality. Where
horizontal steel is used between courses, cement-sand mortar (1:3) should be used
with thickness so as to cover the steel with 6 mm mortar above and below it.
Where vertical steel is used, the surrounding brickwork of 1 X 1 or l½ X 1½ brick
size depending on wall thickness should preferably be built using 1:6 cement-sand
mortar.
o The minimum wall thickness shall be one brick in one storey construction and one
brick in top storey and 1½brick in bottom storeys of up to three storey

constructions. It should also not be less than l/16 of the length of wall between two
consecutive perpendicular walls.
o The height of the building shall be restricted to the following, where each storey
height shall not exceed 3.0 m.
o For Categories A, B and C - three storeys with flat roof; and two storeys plus
attic pitched roof.
o For Category D - two storeys with flat roof; and one storey plus attic for pitched
roof.
o Masonry Construction with Rectangular Masonry Units
 General requirements for construction of masonry walls using rectangular masonry units are:
o Masonry Units
 Well burnt bricks conforming to IS 1077: 1992 or solid concrete blocks
conforming to IS 2185 (Part 1): 1979 and having a crushing strength not
less than 3.5 MPa shall be used. The
strength of masonry unit required shall
depend on the number of storeys and
thickness of walls.
 Squared stone masonry, stone block
masonry or hollow concrete block masonry,
as specified in IS 1597 (Part 2): 1992 of
adequate strength, may also be used.
o Mortar
 Mortars, such as those given in Table – 3 or of equivalent specification,
shall preferably be used for masonry construction for various categories of
buildings.
 Where steel reinforcing bars are provided in masonry the bars shall be
embedded with adequate cover in cement sand mortar not leaner than 1:3
(minimum clear cover 10 mm) or in cement concrete of grade M15
(minimum clear cover 15 mm or bar diameter whichever more), so as to
achieve good bond and corrosion resistance.
o Walls
 Masonry bearing walls built in mortar, as specified in 10.6.3.2 above,
unless rationally designed as reinforced masonry shall not be built of
greater height than 15 m subject to a maximum of four storeys when
measured from the mean ground level to the roof slab or ridge level.

 The bearing walls in both directions shall be straight and symmetrical in
plan as far as possible.
 The wall panels formed between cross walls and floors or roof shall be
checked for their strength in bending as a plate or as a vertical strip
subjected to the earthquake force acting on its own mass. Note — For
panel walls of 200 mm or larger thickness having a storey height not more
than 3.5 metres and laterally supported at the top, this check need not be
exercised.
o Masonry Bond
 For achieving full strength of masonry, the usual bonds specified for
masonry should be followed so that the vertical joints are broken properly
from course to course.
 To obtain full bond between perpendicular walls, it is necessary to make a
slopping (stepped) joint by making the corners first to a height of 600 mm
and then building the wall in between them.
 Otherwise, the toothed joint (as shown in Fig. 10.14) should be made in
both the walls alternatively in lifts of about 450 mm.
 Panel or filler walls in framed buildings shall be properly bonded to
surrounding framing members by means of suitable mortar (as given in
10.6.3.2 above) or connected through dowels.

PART C — (1 x 15 = 15 marks)
16,(a)Determine the natural frequencies of vibration and the ratio of the amplitudes of
motion of mass m1 and m2 for the system shown in figure 16 (a) Here the stiffness is
k1 between the support and mass 1. (15)
figure 16 (a)
Solution:
i). To draw spring mass system:
ii). To draw free body diagram:
iii). To write equation of motion
For mass 1,
̈ = 0
̈ = 0
̈ = 0…… (1)
For mass 2,
̈ + = 0
̈ = 0
̈ - = 0…… (2)
Let us assume the solution of the form,
;
̇ ; ̇
̈ ; ̈

The above equation can be rewritten as,
= 0
= 0
[ ] = 0
[ ] =
[ ]
⁄ = … (3)
- = 0
= 0
[ ] = 0
[ ]
⁄ = … (4)
By equating we get,
[ ]
⁄ = [ ]
⁄
[ ] x [ ] =
{ [ ]} = 0
= 0
The above equation divided into ,
= 0
= 0
( ) = 0
Assume
k1 = 98000 N/m; k2 = 19600 N/m m1=196kg; m2=49kg
( ) = 0
= 0
Take λ = ,
= 0
By solving above equation we get,
= 723.607; = 276.393
;

(Or)
(b) A damped free vibration test is conducted to determine the dynamic properties of a one
storey building. The mass of the building is 10 tonne. Initial displacement of the building is
7.02 mm. Max displacement on the first cycle is 5.3 mm and the period of this displacement
cycle is 1.7s. Determine the effective weight, undamped frequency, logarithmic decrement,
damping ratio, damped frequency and the amplitude after 6 cycles. (15)
To Find:
i.Effective weight
ii.Undamped frequency
iii. Logarithmic decrement
iv. Damping ratio
v.Damped frequency
Given Data:
i. Mass of the building = 10 tonnes =10,000kg
ii. Initial displacement of the building (xo) = 7.02mm
iii. Max. Displacement of the cycle (x1) = 5.3mm
iv. Time period (T) = 1.7 Sec
v. Number of cycle (n) = 6 nos
Solution:
i.Effective weight
⁄
ii. Undamped frequency
Frequency,
⁄ ; ⁄ ;
⁄
⁄
iii. Logarithmic decrement

vi. Damping ratio
vii. Damped frequency
√
√
STAFF INCHARGE HOD PRINCIPAL

Question Paper Nov-Dec-2018.pdf

Recommended

Recommended

More Related Content

What's hot

What's hot (20)

Similar to Question Paper Nov-Dec-2018.pdf

Similar to Question Paper Nov-Dec-2018.pdf (20)

Recently uploaded

Recently uploaded (20)

Question Paper Nov-Dec-2018.pdf