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DEPARTMENT OF CIVIL ENGINEERING Shri Madhwa Vadiraja Institute of Technology and Management (A Unit of Shri Sode Vadiraja Mutt Education Trust ® , Udupi) Vishwothama Nagar, Bantakal – 574 115, Udupi District, Karnataka, INDIA Internal assessment - III Scheme of Evaluation Sub : Structural analysis-II Date: 17 November 2015 Sub Code: 10 CV53 Max Marks : 25 Q. No Solution Marks 1 02 02

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Page 1: EST Scheme

DEPARTMENT OF CIVIL ENGINEERINGShri Madhwa Vadiraja Institute of Technology and Management

(A Unit of Shri Sode Vadiraja Mutt Education Trust®, Udupi)Vishwothama Nagar, Bantakal – 574 115, Udupi District, Karnataka, INDIA

Internal assessment - IIIScheme of Evaluation

Sub : Structural analysis-II Date: 17 November 2015Sub Code: 10 CV53 Max Marks : 25

Q. No Solution Marks

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DEPARTMENT OF CIVIL ENGINEERINGShri Madhwa Vadiraja Institute of Technology and Management

(A Unit of Shri Sode Vadiraja Mutt Education Trust®, Udupi)Vishwothama Nagar, Bantakal – 574 115, Udupi District, Karnataka, INDIA

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DEPARTMENT OF CIVIL ENGINEERINGShri Madhwa Vadiraja Institute of Technology and Management

(A Unit of Shri Sode Vadiraja Mutt Education Trust®, Udupi)Vishwothama Nagar, Bantakal – 574 115, Udupi District, Karnataka, INDIA

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Final moments

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DEPARTMENT OF CIVIL ENGINEERINGShri Madhwa Vadiraja Institute of Technology and Management

(A Unit of Shri Sode Vadiraja Mutt Education Trust®, Udupi)Vishwothama Nagar, Bantakal – 574 115, Udupi District, Karnataka, INDIA

BMD

Free vibration: The vibration which persists in a structure after the force causing the motion has been removed is known as free vibration. No external forces act

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DEPARTMENT OF CIVIL ENGINEERINGShri Madhwa Vadiraja Institute of Technology and Management

(A Unit of Shri Sode Vadiraja Mutt Education Trust®, Udupi)Vishwothama Nagar, Bantakal – 574 115, Udupi District, Karnataka, INDIA

3 on them. It takes place when a system oscillates under the action of forces inherent in the system itself.

Example: Oscillation of a simple pendulum.

Forced vibration: The vibration which is maintained in a structure by steady periodic force acting on the structure is known as forced vibration. When the excitation is oscillatory, the system is forced to vibrate at the excitation frequency.

Damping: Damping is the resistance to the motion of a vibrating body. The vibrations associated with this resistance are known as damped vibrations. It is a phenomenon in which the vibrational energy of the system is gradually reduced or the amplitude of vibration is slowly decreased.

Natural frequency: Generally frequency is the number of cycles per unit time. When no external force acts on the system after giving it an initial displacement the body vibrates. These vibrations are called free vibrations and their frequency is called natural frequency. It is expressed in rad/s or hertz.

Periodic motion: Motion repeated in equal intervals of time. Periodic motion is performed, for example, by a rocking chair, a bouncing ball, a vibrating tuning fork, a swing in motion, the Earth in its orbit around the Sun, and a water wave. In each case the interval of time for a repetition, or cycle, of the motion is called a period, while the number of periods per unit time is called the frequency.

Degree of freedom: It is the number of independent coordinates required to define the position of an object . The position and orientation of a rigid body in space is defined by three components of translation and three components of rotation, which means that it has six degrees of freedom.

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DEPARTMENT OF CIVIL ENGINEERINGShri Madhwa Vadiraja Institute of Technology and Management

(A Unit of Shri Sode Vadiraja Mutt Education Trust®, Udupi)Vishwothama Nagar, Bantakal – 574 115, Udupi District, Karnataka, INDIA

Others resources to be considered Nails Sundries, Tools & Plants Oiling of timber surface 10% contractors profit

a) RCC 1:1.5:3 with graded coarse aggregates of 20mm downsize with 1.2% steel

Solution:To complete 10m3 of RCC work for following resources are required

Materials:Dry material quantity (52% extra) = 15.2m3

Cement – 2.80 m3 or 84 bags Fine aggregates - 4.20 m3

Coarse aggregates (20mm) - 8.40 m3

Steel reinforcement(@1.2%) - 9,420 kgs Binding wire - 2 kgs

Human resource:For Concreting

Head mason = 12

no

Mason = 03 no’s Male Coolie = 12 no’s Female Coolie = 20 no’s Bhisthi = 06 no’s

For Barbending, Binding etc., Bar bender (II Class) - 08 no’s Male coolies - 08 no’s

For Centring & Shuttering (Both erection and dismantling) Carpenter (II Class) - 10 no’s Male Coolies - 10 no’s

Others resources to be considered Sundries, Tools and planks( for Concreting) Concrete mixer Sundries, Tools & Plants (Barbending) Sundries, Tools & Plants (Centring) Nails Timber planks and ballies 1.5% water charges 10% contractors profit

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DEPARTMENT OF CIVIL ENGINEERINGShri Madhwa Vadiraja Institute of Technology and Management

(A Unit of Shri Sode Vadiraja Mutt Education Trust®, Udupi)Vishwothama Nagar, Bantakal – 574 115, Udupi District, Karnataka, INDIA

b) Random Rubble Stone Masonry for foundation & Plinth in CM 1:6

Solution:To complete 10m3 of work for foundation following resources are required

Materials:Dry mortar required (42%) = 4.2 m3

Cement – 0.70 m3 or 21 bags Fine aggregates - 4.20 m3

Stone required - 12.5m3

Human resource:

Head mason = 12

no

Mason = 10 no’s Male Coolie = 08 no’s Female Coolie = 08 no’s

Bhisthi = 112

no’s

Others resources to be considered Sundries, Tools and planks 10% contractors profit

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042 Estimate the cost of earthwork and turfing for the embankment for 350m

length with following data.Formation Level, m

106.8Down gradient in 1 in 100

RL of Ground, m

105.42 104.3 104.80 104.00 102.90 102.00 102.60 101.80

Distance, m

0 50 100 150 200 250 300 350

Formation width is 10m and the side slopes for cutting and Filling are 1.5:1 and 2:1 respectively. Take the cost of earthwork in cutting and filling as `250 per m3

and `280 per m3 respectively. Cost of turfing as `100 per m2

SolutionThe problem can be solved by any one method1. Mid sectional area method2. Mean sectional area method

1. Mid sectional area method For every 100m length there is a down of 1m (1:100), therefore for 50m down of 0.5m

Down gradient in 1 in 100

Formation 106.8 106.3 105.80 105.30 104.80 104.30 103.80 103.30

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DEPARTMENT OF CIVIL ENGINEERINGShri Madhwa Vadiraja Institute of Technology and Management

(A Unit of Shri Sode Vadiraja Mutt Education Trust®, Udupi)Vishwothama Nagar, Bantakal – 574 115, Udupi District, Karnataka, INDIA

Level, mRL of

Ground, m105.42 104.3 104.80 104.00 102.90 102.00 102.60 101.80

Depth of fill, m

1.38 2.00 1.00 1.30 1.90 2.3 1.20 1.50

Distance, m 0 50 100 150 200 250 300 350

B = 10, S = 1.5

Distance Depth, mMean

depth, m Bdm Sdm2 Bdm + Sdm2 Length

Quantity, m3

0 1.38 - - - - - -

50 2.00 1.69 16.90 4.28 21.18 50 1059

100 1.00 1.50 15.00 3.375 18.375 50 918.75

150 1.30 1.15 11.50 1.983 13.483 50 674.15

200 1.90 1.6 16.00 3.84 19.84 50 992

250 2.30 2.1 21.00 6.615 27.615 50 1380.75

300 1.20 1.75 17.50 4.593 22.093 50 1104.65

350 1.50 1.35 13.50 2.734 16.234 50 811.70

Total Quantity in filling, m3 6941

2. Mean Sectional Area method B =10m, S = 1.5

Distance Depth, m Bdm Sdm2 Bdm + Sdm2

Mean Area Length Quantity,

m3

0 1.38 13.80 2.857 16.657 - - -

50 2.00 20.00 6.00 26.00 21.328 50 1066.40

100 1.00 10.00 1.50 11.50 18.75 50 937.50

150 1.30 13.00 2.535 15.535 13.517 50 675.85

200 1.90 19.00 5.415 24.415 19.975 50 998.75

250 2.30 23.00 7.935 30.935 27.675 50 1383.75

300 1.20 12.00 2.16 14.16 22.547 50 1127.35

350 1.50 15.00 3.375 18.375 16.267 50 813.35

Total Quantity in filling, m3 7002.95

Calculation of areas of side slopes for turfing

Distance Mean Depth, dm

Sloping breadthdm√S2+1

LengthArea of both the side slopes

2Ldm√S2+1

0 - - - -

08

08

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DEPARTMENT OF CIVIL ENGINEERINGShri Madhwa Vadiraja Institute of Technology and Management

(A Unit of Shri Sode Vadiraja Mutt Education Trust®, Udupi)Vishwothama Nagar, Bantakal – 574 115, Udupi District, Karnataka, INDIA

50 1.69 3.047 50 304.70

100 1.50 2.704 50 270.40

150 1.15 2.073 50 207.30

200 1.6 2.884 50 288.40

250 2.1 3.786 50 378.60

300 1.75 3.155 50 315.50

350 1.35 2.434 50 243.40

Total Area, m2 2008.30

Abstract of Estimate by applying given rates for turfing and earthwork

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