# How much reinforcement for Footing ?

How do I calculate quantities of reinforcement in a concrete base for a building? by Santosh Kulkarni https://www.quora.com/How-do-I-calculate-quantities-of-reinforcement-in-a-concrete-base-for-a-building/answer/Santosh-Kulkarni-37?share=5b0bd8a8&srid=zM5g

# RCC BEAM FULLY ANALYZED

Dear Friends ,

we are meeting after a long time.

Today , let us analyze a  RCC beam completely as per the drawing.

First , have a look at the RCC drawing for the ground beam.

Now look at the schedule of Beams.

We have randomly selected beam no P7 for the study today.

The beam looks like this :

For finding the cutting lengths of the various bars , we need to know the SPANS.

Look at this sketch.

With this sketch now , we shall calculate the cutting lengths as follows :

# WEIGHT OF ROOFING TRUSS

OBSERVE THE FOLLOWING SKETCH CAREFULLY.

WE NEED TO FIND OUT THE WEIGHT OF 50 SUCH TRUSSES.

STEP 01:

• FIND OUT LENGTH OF ANGLE AC :
• BY USING THE PYTHAGOROUS FORMULA)
• UNKNOWN LENGTH AC =
• SQRT ( 0.6 X0.6 + 1.5 X1.5) = 1.615 M

STEP 02:

• FIND OUT TOTAL LENGTH OF ANGLE USED:
• TOTAL LENGTH: 1.615 +0.6 +1.5+0.3 =4.015 M
• STEP 03:
• WEIGHT OF ANGLE 50 X50 X 5: 3.8 KG/M
• TRUSS WEIGHT = 4.015 X 3.8 =15.257 KG
• TOTAL WEIGHT OF 50 TRUSSES:
•    50 X 15.257 = 762.85KG

# RCC footing : Reinforcement

Exercise :

There is a sketch of a RCC footing.on which we need to work today.

This is a hand drawn sketch of a trapezoidal footing.

The dimensions and the reinforcement details have been given in the sketch.

First thing is to ask several questions and seek the answers :

1. What is the size of the footing in plan ? { 4 m x 3 M }
2. What is the depth of the raft ? { 300 mm }
3. What is the vertical depth of the sloping portion of the footing ? { 900 mm }
4. What is the size of the column in plan ? { 1000 mm x 1000 mm }
5. What is the main steel of the footing ? {16 mm Tor @ 200 mm c/c both ways }
6. What is the reinf. in column ? { 8 bars of 25 mm dia Tor }
7. Column stirrups ? {10mm bars @ 200 mm c/c }
8. Height of the column { 5000 mm }.
9. Cover for  the footing bars : 50mm
10. Cover for the column bars : 40 mm
11. Weight of 16 mm bars : 1.58 Kg/M
12. Weight of 25 mm bars : 3.86 kg/M
13. Weight of 10 mm bars : 0.62 Kg /M.

Thus we know the basic details.

Now let us calculate the weight of the reinforcement .

}1.MAIN BARS OF THE COLUMN: NOS X 25TOR

}TOTAL HEIGHT:

Vertical height + sloping height + Raft height

5000 + 900 + 300 = 6200

}CUTTING LENGTH:

}TOTAL HEIGHT (6200) – BOTTOM COVER (50) -TOP COVER (40) + development length ’’L’’(300)

6200-50-40 + 300 = 6410= 6.41M

WEIGHT:

}25TOR : 8NOS X 6.41M* 3.85KG/M=  197.43KG

}2.STIRRUPS:  10 TOR@200 C/C

}CUTTING LENGTH:

}920*4 = 3680+ Hook length 120 = 3800 = 3.8M

NO OF STIRRUPS: 6110/200+1=32NOS

WEIGHT: 10 TOR:  32 NOS*3.8M*0.62=75.39KG

}

}

}3.RAFT BARS: 16TOR@200C/C BOTH WAYS

}A. LONG SIDE ( 4M):

}CUTTING LENGTH:

}4000-50-50+ development length ( 200+200) =4300= 4.3M

}NO OF BARS:

}(3000/200) +1= 16NOS

WEIGHT:

}16TOR:16NOS*4.3M@1.58KG/M=108.7KG

}

}

}3.RAFT BARS: 16TOR@200C/C BOTH WAYS

}B.SHORT SIDE( 3M):

}CUTTING LENGTH:

}3000-50-50+development length ( 200+200) =3300= 3.3M

}NO OF BARS:

}4000/200+1= 21NOS

}WEIGHT:

}16TOR:21NOS*3.3M @1.58KG/M=109.49KG

}16TOR:TOTAL:108.7KG+109.49KG=218KG

}

}

}

}10 TOR:    76   KG

}16TOR:   218   KG

}25TOR :  198   KG

}

}TOTAL WEIGHT:       492 KG

}

# Structural steel estimation for Roof girder

We have already seen how to perform the basic calculations for finding the structural weight .

Refer to the earlier posts such as :

Today let us see how to find out the weight of a N type of girder.

Please refer to the sketch as below :

The sketch is quite simple and also self-explanatory.

Hence, let us directly calculate the weight as follows :

# RCC Curved beam

One of my students has asked me this question today :

Sir , how to find out the reinforcement for the curved beam ?

So , dear readers , let us solve this question today.

For the sake of understanding , I have assumed the following data for the curved beam.

Curved length of the RCC beam : 8560 MM.

Note :

In case , it is not possible , to measure it from the drawing , due to lack of clarity on the beam curvature , I suggest that first let the beam bottom be fixed at the position and then actually measure the beam curved length at the position. This shall give the most correct beam length at site.

Beam size :

Width :250 MM.

Depth : 650 MM.

Beam top bars : 2 Nos x 12 mm dia.HYSD bars [ Straight ]

Beam Main bars : 4 x 25 mm dia HYSD bars [ Straight ]

Bent up bars : 2 x 25 mm dia HYSD bars

Stirrups : 8 mm dia HYSD bars @ 200 c/c.

We also know that :

Beam cover : 25 MM.

Stirrup hook length : 2 x ( 10 X dia of stirrup bar )

Development length for the top bars : 450 mm.

Development length for the bottom bars : 450 mm.

Now with the above information , let us calculate the beam reinforcement as follows :

1. Top bars : [ 8560- (25-25)+ (450+450) ] = [ 8560 – 50 = 900 ] = 9410 mm = 9.41 M.
2. Bottom straight bars : 9.41M.
3. Bent up bars : ( 8560-50 )+ 2 {0.41( 650-50)} + (450+450) ] = ( 8510 + 2 x 0.41 x 600 )+900  = ( 8510 +492+900 ) = 7902 mm = 7.902 M.
4. Stirrups : Cutting length : 2 { ( 250 -50 ) +( 650 -50 )} + 2 X 10 x8 = 1600 + 160 = 1660 mm =1.660 M                                           No of stirrups : ( 8560/200 ) + 1 = 43.8 say 44 Nos.

Abstract for the weight :

1. Top bars : 2 x 9.41 M x 0.89 Kg/M = 16.75 Kg.
2. Bottom straight bars : 4 x 9.41 M .x 3.86 Kg/M = 145 Kg.
3. Bent up bars : 2 x 7.902 M x 3.86 Kg/M = 61 Kg.
4. Stirrups : Cutting length : 44 x 1.660 M x 0.39 Kg/M =   28.5 Kg.                                                                                                                                                                                                                                                                                                                                                  Total weight :  252 Kg.

# Mild Steel Plate Weight 02

Exercise :

Calculate the weight of a 16 mm thick plate whose dimensions have been measured as 4 m x 2 M.

Solution :

Unit weight of the 16 mm thick plate : 16 x 7.86 = 125.76 KG/SQM

WEIGHT OF THE PLATE:

= AREA X UNIT WEIGHT

=8.0SQM X 125.76 KG/SQM =1006.08KG

=1.006 MT