Family Video

Some Dafination about soil Mechanics

Unit Weight:

Total unit weight: Cumulative weight of the solid particles, water and air in the material per unit volume. Note that the air phase is often assumed to be weightless.

Porosity:

Ratio of the volume of voids (containing air, water, or other fluids) in a soil to the total volume of the soil.

Void ratio:

is the ratio of the volume of voids to the volume of solid particles in a soil. Void ratio is mathematically related to the porosity.

Permeability:

A measure of the ability of water to flow through the soil, expressed in units of velocity.

Compressibility:

The rate of change of volume with effective stress. If the pores are filled with water, then the water must be squeezed out of the pores to allow volumetric compression of the soil; this process is called consolidation.

Shear strength:

The shear stress that will cause shear failure.

Atterberg Limits

Liquid limit, plastic limit, and shrinkage limit. These indices are used for estimation of other engineering properties and for soil classification.

Happy New ear

Happy New year to everybody. Today is the day of wanting. We should not memories the passing year about bed day. We hope coming year will earn very good time. we want to earn success in the coming year. I want success and peaceful life for everybody of the world.
Engr M A Razzak
14.04.2015

test copy

I'm Engr M A Razzak, my son Kader Pial Cowdhury in front of me.

Sample of a Schedule of Electrical works

	Toma Construction & Company Limited						'Annexure-A'
	Tongi-Bhairab Bazaar Double Line Project
	Bill of Quantities
	Bill No-5, Subsection D : Stations & Buildings
Name of works:		Electrical works for Plat form Sheds & Foot Over Bridge.
Sl No.	BOQ No	Name of works		Unit	Quantity	Rate in Taka	Amount in Taka
273	D-3.1(a)	Concealed Point Wiring, Light Point		nr	275	71.86	19761.50
274	D-3.1(b)	Concealed Point Wiring, Fan Point		nr	110	71.86	7904.60
276	D-3.2(a)	Concealed Circuit / Power Point Wiring, 1C-2x2.5 Sq mm Cable with 2.5 sq mm ECC Wire through 16mm diameter PVC Pipe.		Lin m	1800	228.51	411318.00
277	D-3.2(b)	Concealed Circuit / Power Point Wiring, 1C-2x4.0 Sq mm Cable with 4.0 sq mm  ECC Wire through 20mm diameter PVC Pipe.		Lin m	600	228.51	137106.00
278	D-3.2(c)	Concealed Circuit / Power Point Wiring, 1C-2x6.00 Sq mm Cable with 6.0 sq mm  ECC Wire through 25mm diameter PVC Pipe.		Lin m	240	282.40	67776.00
281	D-3.3(c)	1C x 4 x 10 Sq mm Cable with 1C-1x10 sq mm  ECC Wire.		Lin m	600	521.69	313014.00
282	D-3.3(d)	1C x 4 x 16 Sq mm Cable with 1C-1x16 sq mm ECC Wire.		Lin m	600	658.22	394932.00
285	D-3.6(a)	Galvanized Iron (GI) Works, 19mm nominal ID		Lin m	48	697.74	33491.52
290	D-3.7(d)	One	Gang Switch		nr	385	83.36	32093.60
291	D-3.7(e)	Two-GangSwitch		nr	41	121.44	4979.04
292	D-3.7(f)	Three- Gang Switch		nr	39	135.81	5296.59
293	D-3.8	Ceiling Fan, AC Capacitor Type, 1422.4mm Sweep		nr	110	2453.96	269935.60
300	D-3.10(f)	Distribution Board 250V Grade with 6-32A SP MCB.		nr	22	1617.53	35585.66
302	D-3.11(a)	Fency Single Bracket Light Fitting.		nr	44	11016.57	484729.08
303	D-3.11(b)	1-4' x 40 Watt Fluricent Tube.		nr	220	773.19	170101.80
307	D-3.11(f)	40/60 Watt Lamp		nr	66	1833.82	121032.12
308	D-3.11(g)	100 Watt Lamp		nr	44	3334.94	146737.36
		Grand Total Taka					2655794.47
	Prepared By
	Engr M A Razzak
	Site-in-Charge (TBDLRP)

Basic Electrical Engineering

Series circuits:

Series circuits are sometimes called current-coupled or daisy chain-coupled. The current in a series circuit goes through every component in the circuit. Therefore, all of the components in a series connection carry the same current. There is only one path in a series circuit in which the current can flow.
A series circuit's main disadvantage or advantage, depending on its intended role in a product's overall design, is that because there is only one path in which its current can flow, opening or breaking a series circuit at any point causes the entire circuit to "open" or stop operating. For example, if even one of the light bulbs in an older-style string of Christmas tree lights burns out or is removed, the entire string becomes inoperable until the bulb is replaced.

Current

In a series circuit the current is the same for all elements.

Resistors

The total resistance of resistors in series is equal to the sum of their individual resistances:

Electrical conductance presents a reciprocal quantity to resistance. Total conductance of a series circuits of pure resistors, therefore, can be calculated from the following expression:

.

For a special case of two resistors in series, the total conductance is equal to:

 

DEPARTMENT OF CIVIL ENGINEERING 
Program: B. Sc in Civil Engineering

Course Title:

Mechanics, Heat, Thermodynamics, Waves, Oscillations and Physical Optics- Lab

Course code:        PHY 103

Assignment

Experiment No-01                                   Experiment Done Date: 28.01.2015

Name of the Experiment:

To determine the focal length and hence the power of a convex lens by displacement method with the help of an optical bench.

Submitted by:

Name: Mohammed Abdur Razzak

Roll No:  3326,  Registration No: WUB/10/13/60/3326, 

Batch No: 60/C, Semester No: Fourth, Year: Second.

Submitted To:

Name:  Md. Shohel Rana

 Designation:  Lecturer in physics, Basic science Division (WUB)                 Date of Submission:

Experiment No.:04

Experiment Name: Determine the focal length and hence the power of a convex lens by displacement method with the help of an optical bench.

Theory: If the object and the image screen be so placed on an optical bench that the distance between them is greater than four times the focal length (f) of a given convex lens, then there will be two different positions of the lens which for which an equal sharp image will be obtained on the image screen.

                                                                                                                     

The fig represent respectively the positions of the object and the image screen and the two different positions of the lens for which an equally sharp image is obtained.

Let the distance Ol =D and L₁ – L₂ = x.

From the lens equation , we have

Or,   (since u + v =D)

Applying sign convention, u is negative.

Or,

Or, u² – ud =df = 0

Solving the above equation which is quartic, we have two values of u corresponding to the two positions of the lens. These are

u₁ =  -  position L₁ of the lens

And u₂ =  +  position L₂ of the lens

Then x = L₁  L₂ = u₁  u₂ =

Or   x² = D² – 4Df

Or   f =  ……………….. (1)

Where D is the distance between the object and the image and must be greater than 4f and x is the distance between two different positions of the lens.

The power P of the lens s as usual given by the relation,

P =  diopters

Results:

Table I

No. of obs.	Position of				Displacement of Lens X = L1 – L2 (cm)	Apparent distance between object and image D’ = O	Corrected distance between object and image D = D’ +
	Object (O)	Image (I)	Lens at
			L1	L2
1	0	45	16	28.2	12.20	45
2	0	50	14.4	34.4	20.00	50
3	0	55	14	40.30	26.30	55
4	0	60	12.4	46.00	33.60	60
5	0	65	12.1	51.00	38.90	65

Table II:

No. of obs.	Lens displacement (x) from Tab. II	Corrected Distance (d) from Tab. II	Focal length	Mean focal length (f) cm	Power P =
1	12.20	45	10.42	10.45	9.56
2	20.00	50	10.50
3	26.30	55	10.60
4	33.60	60	10.30
5	38.90	65	10.43