Scoffolding - Types & its Details

Scaffolding work is explained simply in below images

How many bricks required for 1 cu.m wall ???

1.     How many bricks required for 1 cu.m wall of 1^1/2 brick thickness ???

ANSWER:

 

                                       For a 1⅟₂   brick thickiness wall (i.e. 30cm nominal thickness). Therefore nominal volume of wall of 20m length , 1m  height & 0.3m thickness = 20*5*.3 = 30 cu.m.

                          Normally mortar joint will be less than 1cm, but assuming mortar joint as   1cm thickness, then actual thickness of wall will be = 29cm. therefore , actual thickness = 20*5*.29 = 29 cu.m.

      Nominal Size of standard brick =  20cm*10cm*10cm.

 

       Number of std bricks = 29/(.20*.10*.10) =  14500

       Then,

           Number of bricks per cu.m(nominal) = 14500/30 =483.3333

                                                                                =483 bricks.

      Considering 5% breakages, wasteges,etc (assume)

                    So,Number of bricks required for 1 cu.m = 500 bricks

        RESULT:

        Therefore, number of bricks required for 1 cu.m = 500 bricks .

DESIGN OF A CONCRETE APPROACH ROAD AT EYYAKUNAM VILLAGE --- PROJECT REPORT

   Project available @ below link.....

                              This design project titled “Design of cement concrete approach road at Eyyakunam village’’ deals with the pavements in general, its types and design factors taking a case study of a road leading to a small village between Gingee and Tiruvannamalai. The concentration of the project report is keened on the area of design of rigid pavement which is the best suited pavement design for the road in consideration. The design of rigid pavement for the proposed road is illustrated and designed for the soil condition and traffic condition of the approach road to Eyyakunam village. Codal recommendations provided by the Indian Road Congress, “Guidelines for the design of flexible pavements for highways” (IRC 37-2001) are strictly followed in preparing this design report. Codal recommendations provided by the Indian Road Congress, “Guidelines for the design of plain jointed rigid pavements for highways” (IRC 58-2002) are strictly followed in preparing this design report.

The next part of the report, sandwiches the above manual pavement design with the help of a flowchart methodology. Computerisation of this process will reduce the consumption of design time, complexity in manual designing and increase the reliability.

For more details..............

Click here to download.

CHALLENGING TEST FOR CIVIL ENGINEERS & CHANCE TO WIN FREE RECHARGES

HI CIVILERS......

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RESULTS WILL BE ANNOUNCED ON APRIL 2ND 2017.....


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Important Unit conversion for Civil Engineers

Unit conversion for Civil Engineers:

         Mostly useful for site Engineers...

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HOW TO CALCULATE QUANTITIES OF CEMENT, SAND AND AGGREGATE FOR NOMINAL CONCRETE MIX (1:2:4)?

       Mix design is a process of determining the right quality materials and their relative proportions to prepare concrete of desired properties like workability, strength, setting time and durability.

While following a mix design is advised to optimise the material consumption, it is not possible at site to always come up with Mix design. Nominal mix concrete is prepared by approximate proportioning of cement, sand and aggregate to obtain target compressive strength.

Mix ratio of concrete defines the ratio of cement sand and aggregate by volume in that order. So a mix ratio of 1:2:4 represents Cement: Sand: Aggregate – 1:2:4 (by Volume)

Material requirement for producing 1 Cum of Nominal Concrete Mix

The following are the materials required to produce 1 Cum of Concrete of a given Nominal Mix Proportion.

Method-1: 

 
DLBD method to determine material requirement for Nominal Concrete Mix (M20 – 1:2:4)

The DLBD (Dry Loose Bulk Densities) method is an accurate method to calculate cement, sand and aggregate for a given nominal mix concrete. This gives accurate results as it takes into account the Dry Loose Bulk Densities of materials like Sand and Aggregate which varies based on the local source of the material

For calculation, We consider a nominal concrete mix proportion of 1:2:4 (~M20).

Step-1:
Calculated Volume of materials required.:-

01 bag (50 kg) of cement = 35 litres or 0.035 cubic meter (cum)

Since we know the ratio of cement to sand (1:2) and cement to aggregate (1:4)

Volume of Sand required would be = 0.035*2 = 0.07 cubic meter (cum)

Volume of Aggregate required would be = 0.035*4 = 0.14 cubic meter (cum)

Step-2:
Convert Volume requirement to weights:-

To convert Sand volume into weight we assume,  we need the dry loose bulk density (DLBD). This density for practical purposes has to be determined at site for arriving at the exact quantities. We can also assume the following dry loose bulk densities for calculation.

DLBD of Sand = 1600 kgs/cum

DLBD of Aggregate = 1450 Kgs/Cum

So, Sand required = 0.07*1600 = 112 kgs

and Aggregate required = 0.14*1450 =203 kgs

Considering water/cement (W/C) ratio of 0.55

We can also arrive at the Water required = 50*0.55 = 27.5 kg

So, One bag of cement (50 Kgs) has to be mixed with 112 kgs of Sand, 203 Kgs of aggregate and 27.5 kgs of water to produce M20 grade concrete.

Step-3:
Calculate Material requirement for producing 1 cum Concrete

From the above calculation, we have already got the weights of individual ingredients in concrete.

So, the weight of concrete produced with 1 Bag of cement (50 Kgs) =50 kg + 140 kg + 203 kg + 27.5 kg = 420.5 kg say 420 kgs

Considering concrete density = 2400 kg/cum,

One bag of cement and other ingredients can produce = 420/2400 = 0.175 Cum of concrete (1:2:4)

01 bag cement yield = 0.175 cum concrete with a proportion of 1:2:4

01 cum of concrete will require

Cement required = 1/0.175 = 5.72 Bags

Sand required = 140/0.175 = 800 Kgs

Aggregate required = 203/0.175 = 1160 kgs

Method-2:

 
Empirical method to determine material requirement for Nominal Concrete Mix:-

Although empirical method is easy to use in determining the materials requirement for Nominal Concrete mix, it sometimes doesnt give accurate results as it doesn’t take into factor the local variations in the materials.

Let’s design M20 grade concrete. Ratio for M20 concrete is 1 : 2 : 4

Step-1:
Calculate the Volumes of material required in 1 Cum concrete

The dry volume of concrete mixture is always greater than the wet volume. The ratio of dry volume to the wet volume of concrete is 1.54.

So 1.54 Cum of dry materials (cement, sand and aggregate) is required to produce 1 Cum of concrete

Volume of Cement required = 1/(1+2+4) X 1.54 = 1/7 X 1.54 = 0.22 Cum

Volume of Sand required = 2/7 X 1.54 = 0.44 Cum or 15.53 cft

Volume of Aggregate required = 4/7 X 1.54 = 0.88 Cum or 31.05 cft

Note: 1 cubic meter = 35.29 cubic feet.

Step-2:
Calculate the weights of materials required in 1 Cum concrete:-

Density of Cement (loose) = 1440 kgs/cum

So weight of cement required = 1440 X 0.22 = 316.2 Kgs or 6.32 bags

Density of Sand = 1600 Kgs/cum

Weight of Sand required  = 1600 X 0.44 = 704 kgs

Density of Aggregate = 1450 kgs/cum

Weight of aggregate required = 1450 X 0.88 = 1,276 Kgs

Types of beam based on Reinforcements

SINGLY REINFORCED SECTION :

When the beam us loaded it can suffer through two different type of beam(mode of failure):

1.Ductile Failure &

2.Brittle Failure

Now, in case of a brittle failure of beam when a beam gets over-stressed, then till the point of over-stressing any major cracks aren’t developed in beam, but as soon the point is passed, suddenly the beam fails.Well thus doesn’t seem good h..!! Hence we always design the beam to fail in the ductile fashion, consequently giving us more time to notice that the beam is over-stressed and it’s the time that we should do some repairs or evacuate the building.

Now, how do we make sure that the beam will fail in ductile manner. This can only possible when the strength of the steel will be less than compression strength of the concrete. This means that when the ultimate loads are reached, as the total strength of the steel will be  pretty less than the strength of concrete, steel will try to strain more than what the concrete will experience strain in compression. Now we all know that steel is ductile and able to handle higher strains without any kind of failure. So this is how we can get a ductile response of beam. In this case, Concrete is all in under control while steel is straining itself.

But suppose if  there are certain restrictions like we cannot use a section greater than say 25″ x 25″ & we have to deal with very high moments. Now this high moment will tend to increase the tension steel demand & consequently a bigger concrete block will be required in compression resulting in pushing the neutral axis further down. Now this is an important part. One might think that so what if neutral axis go further down, I have the whole concrete beam which can take the compression. But wait..!!

As a neutral axis starts shifting down, the strains in the extreme fiber starts to increase. And this will be our concern in case of the compression fiber. Now concrete is brittle, so at start/first it will not show that much impact. But this high value of the strain will cause concrete to crush itself which will lead to the ultimate failure and this will be sudden.  That’s why we can’t put more steel than a certain amount.

So, what decides this limit of steel ? Well, it’s all experimental based approach & to an extent we can prove it mathematically too. Balancing tension & compressive forces, take the depth of a compression block, now draw the strains and look if the strain in the concrete is more than the allowable strain. But in case of designing by us, then some codes specify the maximum r/f ratio in beam whereas some codes specify the maximum allowable neutral axis depth that concrete beam can achieve. And if the desing is under this, thenductile failure in a beam will occur and we are safe.

DOUBLY REINFORCED SECTION :

Now as we mentioned before that we are restricted to a 24″ x 24″ section & we have pretty heavy moment & we are exceeding the limits mentioned in the code. So now the only option is to provide thr compression reinforcement & add some tension steel. Now the extent to which we are adding the compression reinforcement will tell us whether the beam is brittle or ductile.

Suppose the limit for tension reinforcement for singly reinforced beam is X, & to resist this moment we have to add extra 0.5X of tension reinforcements. We also decide to include 0.8X of compression reinforcement. Hence,  now we have 1.5X(1X+.5X) of tension reinforcement & 0.8X of compression reinforcement provided along with compression stress block. So now what happens is that this 0.8X of compression r/f will tend to balance the effect of 0.5X of tension r/f. As we have added 0.8X of compression r/f, the beam doesn’t experience a very high compression strain compared to tension strain. Thus the compression bars will be under lower stress than its yield point. Thus to balance a fully stressed 0.5X of tension r/f we will need a higher amount of compression r/f . Now, the concrete block is responsible to resist the X amount of the tension r/f which is a limit for ductile behavior. Hence in this case the doubly reinforced beam will act as a ductile beam and so we can say that it is an under-reinforced beam/section. All good..!!

But suppose we decide to add only 0.4X of extra compression r/f. Well This will counter max of 0.4X of tension r/f in worst case scenario(it may be lower). So now, the plain concrete stress block will be responsible to resist 1.1X [1X + (.5X-.4X) ]of tension r/f and this is above the limit resulting in brittle failure. So in this case, even though the beam is doubly reinforced, it will experience a brittle failure i.e. it’s an over-reinforced section.