District Rate 2079/080 Kathmandu(आर्थिक वर्ष २०७९।०८० को जिल्ला दररेट काठमाडौँ )

आर्थिक वर्ष २०७९।०८० को जिल्ला दररेट कायम गरिएको बारे

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District Rate 2078/079 Kathmandu(आर्थिक वर्ष २०७९।०८०को जिल्ला दररेट काठमाडौँ )

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Structure Engineering Books Collection

1. Strength of Materials by R.K Bansal

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District Rate 2078/079 Surkhet (आर्थिक वर्ष २०७८।०७९को जिल्ला दररेट सुर्खेत)

 

आर्थिक वर्ष २०७८।०७९ को जिल्ला दररेट कायम गरिएको बारे

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District Rate 2078/079 Surkhet(आर्थिक वर्ष २०७८।०७९को जिल्ला दररेट सुर्खेत)

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District Rate 2078/079 Kathmandu(आर्थिक वर्ष २०७८।०७९को जिल्ला दररेट काठमाडौँ )

 

आर्थिक वर्ष २०७८।०७९  को जिल्ला दररेट कायम गरिएको बारे

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District Rate 2078/079 Kathmandu(आर्थिक वर्ष २०७८।०७९को जिल्ला दररेट काठमाडौँ )

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Centre of Mass and Centre of Stiffness In Etabs

 

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Defining Response Spectrum In Etabs Using NBC105 :2020

 

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Torsional irregularity check in building using NBC 105:2077

 

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Example of Base Shear Calculation using NBC 105:2077

 

A five story reinforce concrete building is to be designed as MRFS and is located in Kathmandu having very soft soil. Determine the seismic base shear according to NBC105:2077                 

L.L=3kN/m² Floor Finish= 1kN/m² Roof Live load=1.5kN/m² importance factor =1.25 M25 concrete Fe500 steel

Slab Thickness: 125 mm Beam: 300*450 Column: 450*450 Story Height =3.2m

                  

Solution:

Calculation of Total DL and LL

Weight of Slab: 5*20*12*0.125*25

                             =3750kN

Weight of beam= (20*4+12*5)*5*0.3*0.45*25

                             =2362.5kN

Weight of Column=0.45*0.45*20*(3.2-0.45)*5*25

                                =1392.1875kN

Floor Finish =20*12*5*1

                    =1200kN

Live load =20*12*3*4

                =2880kN

Roof Live Load =1.5*20*12

                            =360kN

Total Dead load = weight of (Slab + beam + column+ floor Finish)

                           =3750+2362.5+1392.1875+1200

                            =8704.6875kN

Total Live Load = 2880+360

                          =3240kN

 

Seismic weight

The seismic weight at each level, Wi shall be taken as the sum of the dead loads and the factored seismic live loads between the mid-heights of adjacent stories.

The seismic live load is taken as 0.3LL

Floor	Column(kN)	Beam(kN)	Slab(kN)	Floor Finish(kN)	Live Load(kN)	Seismic Weight(DL+0.3LL)
1floor	278.4375	472.5	750	240	720	1965.9375
2nd Floor	278.4375	472.5	750	240	720	1965.9375
3rd Floor	278.4375	472.5	750	240	720	1965.9375
4th Floor	278.4375	472.5	750	240	720	1965.9375
Top Level	139.21875	472.5	750	240	0	1601.7185
Sum=						=9465.4685kN

 

Seismic Weight =9465.4685kN

The approximate fundamental period of vibration, T1, in seconds is determined from following empirical equation

T1 = k_t *H ^¾

  =0.075*16^0.75

       =0.6sec                                                                                                                                        

The approximate fundamental time period calculated using empirical equation shall be increased by a factor of 1.25.

T=1.25*0.6

     =0.75sec

For Kathmandu Very Soft Soil (Type D soil)

Z=0.35 Importance Factor I=1.25

For T=0.75 and D Type soil the

The Spectral Shape Factor C_h(T)=2.25

Elastic Site Spectra for horizontal loading is given by C(T)=C_h(t)*Z*I

                                                                                                 =2.25*0.35*1.25

                                                                                                 =0.984375

Reinforced Concrete Moment Resisting Frame

Ductility Factor(Rμ)= 4               Over strength Factor Ultimate limit State(Ωu)= 1.5

For the ultimate limit state, the horizontal base shear co-efficient shall as given by:

Cd(𝑇𝑖)=C(𝑇𝑖)/Rμ x Ωu

                =0.9843/4*1.5

             =0.1640

Base Shear Force= base shear co-efficient*seismic Weight

                            =0.1640*9465.4685kN

                              V=1552.34kN

VERTICAL DISTRIBUTION OF SEISMIC FORCES

The lateral seismic force (Fi) induced at each level ‘i’ shall be calculated as:

Fi=(Wihi^k/ΣWihi^k )x V

K=1.125 for 0.75 Sec

Floor	Height	Weight	Wi*hi1.125	Seismic Force	Shear Force
Top Level	16	1601.7185	36242.75	472.6468	472.6468
4th Floor	12.8	1965.9375	34608.35	451.3324	923.9792
3rd Floor	9.6	1965.9375	25039.45	326.543	1250.522
2nd Floor	6.4	1965.9375	15868	206.9368	1457.459
1st Floor	3.2	1965.9375	7275.51	94.88095	1552.34
			Sum=119034.1

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Sieve analysis of soil

 

 

Sieve analysis of coarse and fine grained soil                           

Objective and Scope: The objective of the experiment to determine the grain size distribution of coarse grain soil by sieving.

Material and equipment

·         Balance accurate to 1gm

·         Set of is sieves

·         Thermostatically controlled oven

·         Water tight trays

·         Mechanical sieve shaker

·         Riffler

Theory:

In order to classify a soil for engineering purpose, one needs to knows the distributions of the size of grains in soil sample. Sieve analysis is a method to determine the grain size distribution of soils. Sieves are made of woven wires with square openings.

The soil is sieved through the set of sieve. The material is retained on the different sieve is determined. The percentage of material retained in sieve is given by

                                  

Where, M_n=Mass of soil retained on sieve “

           M=Total Mass of soil sample 

 

The results of the mechanical analysis are plotted to get a particle size distribution curve with the percentage finer N as the ordinate and the particle diameter as the abscissa, the dia1neter being plotted on a logarithmic scale. Figure shows some typical curves for various soils. A particle size distribution curve gives us an idea about the type and gradation of the soil. A curve situated higher up or to the left represents a relatively fine grained soil while a curve situated to the right represents a coarse grained soil.

A soil sample may be either well graded or poorly graded (uniformly graded). A soil is said to be well graded when it has good representation of particles of all sizes. On the other hand, a soil is said to be poorly graded if it has an excess of certain particles and deficiency of other, or if it has 1nost of the particles of about the sa1ne size; in the latter case it is known as a uniformly graded soil. Thus, soil A (Fig) is well graded while soil B is uniformly graded. A curve with a flat portion represents a soil in which some intermediate size particles are 1nissing (soil E in Fig.). Such a soil is known as gap graded or skip graded.

For coarse grained soil, certain particle sizes such as D₁₀ D₃₀ and D₆₀ are important. The D₁₀ represents a size, in mm such that 10% of the particle les are finer than this size. Similarly, the soil particle less finer than D₆₀ size are 60 per cent of the total mass of the sample. The size D 10 is sometimes called the effective size or effective diameter. The unifbm1ity coefficient Cu (or coefficient of uniformity) is a measure of particle size range and is given by the ratio of D₆₀ and D₁₀ sizes:

                                   

 

Similarly the shape of particles size curve is represented by the coefficient of curvature is given by ,

 

                       

 

For uniformly graded soil C_U is equal to unity. For a well graded soil C_C must be equal to 1to 3 in additional C_U must be greater than 4 for gravel and 6 for sands (USBR1960)

Procedure:

For coarse sieve analysis

·        Take the required quantity of sample. Sieve it through a 4.75mm IS sieve. Take the soil fraction retained on 4.75mm IS sieve for the coarse sieve analysis and that passing through the sieve for the fine sieve analysis.

·        Sieve the sample through the set of coarse sieve in mechanical sieve.

·        Determine the mass of the materials retained in each sieve.

·        Calculate the percentage retained through each sieve on the basis of total mass of the sample taken.

·        Determine the percentage passing through each sieve.

For Fine sieve analysis

·        Take the portion of the soil sample passing 4.75mm IS sieve. Oven dry it at 105 to 110⁰C. weight it to 0.1% of the soil mass.

·        Sieve the sample through the set of coarse sieve in mechanical sieve.

·        Take the material retained on various sieves in a mortar. Rub it with rubber pestle, but do not try to break individual particles.

·        Revise the materials through the nest of sieve.(minimum 1o minute od shaking)

·        Collect the soil fraction retained on each sieve in a separate container. Take the mass

·        Determine the percentage retained, cumulative percentage retained and percentage finer based on the total mass of sample taken.

Observation and calculations:

Total mass of Dry soil= 400gm

Mass of soil retained on 4.75mm sieve= 200gm

Mass of soil passing through 4.75mm sieve= 200gm

S.N
	IS sieve(mm)	Mass of soil retained (gm)	Percentage retained (%)	Cumulative % retained	% finer
Coarse fraction
1	100	-
2	80	-
3	40	-
4	20	30gm	7.50	7.50	92.50
5	10	62gm	15.50	23.00	77.00
6	4.75	108gm	27	50.00	50.00
Fine fraction
7	2	30.5gm	7.62	57.62	42.38
8	1	24.0gm	6.0	63.62	36.38
9	600µ	17.5gm	4.38	68.00	32.00
10	425µ	16.0gm	4.0	72.00	28.00
11	300µ	14.0gm	3.50	75.50	24.50
12	212µ	18.0gm	4.50	80.00	20.00
13	125µ	22.0gm	5.50	85.50	14.50
14	75µ	36.0gm	9.0	94.50	5.50
15	Pan	22.0gm	5.50	100.00

 

 

 

Result:

Percentage finer given in the last column can be used to plot the particle size distribution curve with particle size as abscissa on log scale and the percentage finer and ordinate. [Note. If the fine fraction contains as appreciable amount of clay particles, the wet sieve analysis is required]

Conclusion

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