Define air pollution. Enlist natural and manmade air pollution

 Define air pollution. Enlist natural and manmade air pollution. What are the effects of air pollution on human, plant and materials ?

Solution:

Air pollution: When the air gets contaminates with such substances so that the resulting air poses life threat to human and other living organisms then this is called as air pollution. These substance which pollute the air are called as air pollutants.

Natural air pollution :

 Natural air pollution is caused due to:

(i) Duststroms

(ii) Volcanoes spreading ash and other toxic gases in the environment

Man made air pollution: Manmade air pollution is caused due to:

(i) Emission from automobiles

(ii) Industrial emission

(iii) Dust originating from construction sites

(iv) Open air burning of agricultural wastes and other refuse

(v) Excessive use CFC as refrigerant

(vi) Burning of coal in thermal power plants

Effects of air pollution : 

Air pollution has many hazardous effects on human, plants and other living beings. These are enumerated below:

(i) Sulphur dioxide causes irritation when inhaled and has an adverse effect on the mucous membrane. It increases the breathing rate and causes oxygen deficiency.

(ii) Hydrogen sulphide is a foul smelling gas and cause headache, sleeplessness and conjunctivitis along with pain in the eye. Higher concentration of this pollutant may lead to blockage of oxygen transfer in humans.

(iii) Carbon monoxide: It results due to incomplete combustion of coal. When inhaled, it replaces oxygen in hemoglobin and forms carboxy – hemoglobin (CO.Hb) and thus human tissues gets devoid of oxygen which may lead to death.

(iv) Oxides of nitrogen: There area total of five oxides of nitrogen viz N2O,NO,NO2,NO3 and N2 O5. Among these, only the NO and NO2 are injurious to human health. These cause nasal and eye irritation along with respiratory disorder.

(v) Lead: lead cause irritation to mucous membranes of nose, throat and lungs. Lead has fatal effect on gastrointestinal tract, kidney and lever. It retards mental growth.

(vi) Fluorides: hydrogen fluoride and other fluorides cause fluorosis in cattle.Certain varieties of fluorides are highly irritant and corrosive in nature.

(vii) Benzene: Benzene and other aromatic hydrocarbons cause canser.

Classify the solid wastes giving suitable example

 Classify the solid wastes, giving suitable example for each of them. Also explain the different methods of disposal of solid wastes.

Solution:

Refuse represents the dry wastes or solid wastes of the society except human excreta and sullage. It includes garbage, ashes, rubbish, dust, etc.

(i) Garbage includes all sorts of putrescible organic wastes obtained from kitchens, hotels, restautants, etc. All waste food

articles, vegetable peelings, fruit peelings, etc., are thus included in this term.

(ii) Ashes denote the incombustible waste products from heaths and furnaces, and houses or industries.

(iii) Rubbish includes all non – putrescible wastes except ashes. It, thus includes all combustible and non- combustible wastes such as rage, paper pieces, broken pieces of glass and furniture, card boards, broken crockery, etc.

Besides the technical classification based on the type of wastes, the refuse may also be classified, depending on its source, as: (i) house refuse; (ii) street refuse and (iii) Trade refuse.

Collection of refuse: The refuse is generally collected in individual houses in small containers and from there, it is collected by sweepers in small hand driven lorries/carts and then dumped into the masonry chambers constructed along roadsides by municipalities. The refuse is finally carted away by municipal trucks, for further disposal during some day time.

Disposal of refuse:

The refuse can be disposed of by various methods, such as

(a) Sanitary land filing: In this method of refuse disposal, refuse is carried and dumped into the low lying area under an engineered operation, designed and operated according to the acceptable standards, as not to cause any nuisance or hazards to public health or safety. Area method and trench method are two methods of land filing which are usually adopted.

(b) Burning or incineration: Burning of refuse at high temperatures in furnaces called incinerators is quite a sanitary method of refuse disposal and is widely adopted in USA where the refuse is of high calorific value and hence quite suitable for burning. The heat produced during burning of the refuse is used in the form of steam power for running turbines to generate electricity.

(c) Barging it out into the sea: This method may be used to dispose of refuse by throwing it away into the sea, after carrying it at reasonable distance from the coast. The sea depth at such disposal point should not be less than 30m or so, and the direction of the currents should be as not to bring it back towards the shore.

(d) Pulverization: In this method refuse is pulverized in grinding machine, so as to reduce its volume and to change its physical character. The grinded or pulverized refuse becomes practically odorless and unattractive to this insects.

(e) Composting: It is biological method of decomposing solid wastes. This decomposition can be effected either under aerobic or anaerobic conditions or both. The final end product is manure called compost or human, which is in great demand as fertilizer for farms.

numerate the situations in which doubly reinforced concrete

 Enumerate the situations in which doubly reinforced concrete beams become necessary. What is the role of compression steel?

Solution :-

When the bending moment required to be resisted is more than the moment of resistance of a balanced section of singly reinforced beam of given size, there are two alternatives:

(1) To use an over-reinforced section

(ii) To use doubly reinforced section. An over reinforced section is always uneconomical and also undesirable because of sudden failure probability. Also the increase in the moment of resistance is not in proportion to the increase in the area of tensile reinforcement. The reason behind this is that the concrete, having reached maximum allowable stress, cannot take more additional load without adding compression steel. The other alternative is to provide reinforcement in the compression side of the beam and thus to increase the moment of resistance of the beam beyond that of a balanced section.

Doubly reinforced sections are also useful in following situations :-

(1) Where the members are subjected to probable reversal of external loads and thereby the bending moment in the section reverses, such as in concrete piles etc. 

(2)When the members are subjected to loading, eccentric to either side of the axis, such as in columns subjected to wind loads.

(3) When the members are subjected to accidental lateral loads, shock or impact.

The steel reinforcement provided in the compression zone is subjected to compressive stress. However, concrete undergoes creep strains due to continued compressive stress, with the result that the strain in concrete goes on increasing with time. This increases compressive strain in steel in addition to creep strain in compressive steel. Thus the total compressive strain in compressive steel will be much greater than the strain in surrounding concrete due to flexure alone. Thus, compressive steel takes up all the additional compressive stresses beyond the permissible compressive stress for concrete making the section safe against failure in flexure.

What are the advantages and disadvantages of welded joints

 What are the advantages and disadvantages of welded joints?

Solution :-

Advantages of welded joints :-

1. Welded joints are more economical than riveted or bolted joints because splice plates and rivets are not required.

2. Welded joints are more rigid than riveted joints. The cover plates, connecting angles etc. in riveted joints make the joint more flexible.

3. It is because of welding that now it is very easy to joint tubular section.

4. Due to fusion of two metal pieces, the structure obtained is continuous. thus welded joints give better architectural appearance.

5. Welding process does not make large sound as compared to riveted joints and thus less noise pollution is there.

6. Because of the absence of splice plates, connecting angles etc, the drawing detailing also gets reduced thereby saving in cost and time.

7. The efficiency of welded joint is more than that of riveted joint.

Disadvantages of welded joints ;-

1. More skilled personnel are required for making welds in the metal.

2. The inspection of welded joint is difficult and expensive.

3. The heat generated during the welding process may distort the connecting members.

4. Welded joints are more prone for brittle failure as compared to other joints.

The torsional resistance of a shaft is directly proportional to

 Pick up the incorrect statement from the following: The torsional resistance of a shaft is directly proportional to

A) modulus of rigidity

B) angle of twist

C) reciprocal of the length of the shaft

D) moment of inertia of the shaft section

Solution-

Using torsion equation ,

TJ=GθL=τR$\frac{\mathrm{T}}{\mathrm{J}}=\frac{\mathrm{G}\theta }{\mathrm{L}}=\frac{\tau }{\mathrm{R}}$

$\frac{\mathrm{<br>}}{}$

Where,

T = torsion applied, J = Polar moment of inertia, G = modulus of rigidity,

θ = Twist of cross-section, L = Length of the shaft,

τ = Maximum shear stress due to torsion and R = Radial distance from the centre of the shaft

From the above equation, torsional resistance of a shaft is directly proportional to the modulus of rigidity, angle of twist, maximum shear stress due to torsion, polar moment of inertia while inversely proportional to the length of the shaft and radial distance from the centre of the shaft.

Thus, it does not depends on the moment of inertia of the shaft section but depends on the polar moment of inertia of the shaft.

∴ The torsional resistance of a shaft is directly proportional to the modulus of rigidity, angle of twist, reciprocal of the length of the shaft.

If the angle of shearing resistance is 60° then the product of active earth pressure

If the angle of shearing resistance is 60°, then the product of active earth pressure coefficient and passive earth pressure coefficient will be

A) 0

B) 1/3

C) 1

D) 3

Coefficient of Active Earth Pressure Coefficient :-

Ka=1−sin(ϕ)1+sin(ϕ)=tan2(45−ϕ2)$K_a=\frac{1-\sin \left(\varphi \right)}{1+\sin \left(\varphi \right)}={\tan }^2\left(45-\frac{\varphi }{2}\right)$

$<br>$

Coefficient of Passive Earth Pressure Coefficient :-

Kp=1+sin(ϕ)1−sin(ϕ)=tan2(45+ϕ2)$K_p=\frac{1+\sin \left(\varphi \right)}{1-\sin \left(\varphi \right)}={\tan }^2\left(45+\frac{\varphi }{2}\right)$

Calculations:

Point to be remembered:

For any value of shearing angle, the product of active earth pressure coefficient and passive earth pressure coefficient will always equal to 1.

Ka×Kp=1−sin(ϕ)1+sin(ϕ)×1+sin(ϕ)1−sin(ϕ)=1$K_a\times K_p=\frac{1-\sin \left(\varphi \right)}{1+\sin \left(\varphi \right)}\times \frac{1+\sin \left(\varphi \right)}{1-\sin \left(\varphi \right)}=1$

Alternate Solution:

For ϕ = 60°

Ka=tan2(45−602)=0.0718$K_a={\tan }^2\left(45-\frac{60}{2}\right)=0.0718$

Ka=tan2(45+602)=13.928$K_a={\tan }^2\left(45+\frac{60}{2}\right)=13.928$

Therefore,

Ka × Kp = 0.0718 × 13.928 = 1