Basic reinforcement knowledge for site engineers
Basic reinforcement knowledge for site engineers
Re: Deduction in length of reinforcement steel for bar bending
Answer
# 1 For 45 degree bend deduct 1d.
For 90 degree bend deduct 2d.
d= dia of bar.
Re: how can we calculate cutting length of a column ring (300 x 600)
Answer
# 1 Cutting Length of a column Formula =
2(L+B)+20d-2x10d
So,
L=600-80=520mm
B=300-80=220mm
Cutting Length = 2 (520+220) + 10x8
= 2(740)+80
= 1480+80=1560 mm
if we Assume that the dia of ring is 8mm.
Things Site Engineers Must Know About Reinforcement and Steel Bars
Clear cover to main reinforcement in
Footings : 50 mm
Raft foundation Top : 50 mm
Raft foundation Bottom/ sides : 75 mm
Strap Beam : 50 mm
Grade Slab : 20 mm
Column : 40 mm (d>12mm) 25 mm (d= 12mm)
Shear Wall : 25 mm
Beams : 25 mm
Slabs : 15 mm or not less than diameter of the bar.
Flat Slab : 20 mm
Staircase : 15 mm
Retaining Wall on Earth : 20/ 25 mm
Water retaining structures : 20 / 30 mm
Sunshade (Chajja) : 25 mm
Footings : 50 mm
Raft foundation Top : 50 mm
Raft foundation Bottom/ sides : 75 mm
Strap Beam : 50 mm
Grade Slab : 20 mm
Column : 40 mm (d>12mm) 25 mm (d= 12mm)
Shear Wall : 25 mm
Beams : 25 mm
Slabs : 15 mm or not less than diameter of the bar.
Flat Slab : 20 mm
Staircase : 15 mm
Retaining Wall on Earth : 20/ 25 mm
Water retaining structures : 20 / 30 mm
Sunshade (Chajja) : 25 mm
Hook for stirrups is 9D for one side
No. of stirrups = (clear span/Spanning) + 1
For Cantilever anchorage length for main steel is 69D
“L” for column main rod in footing is minimum of 300mm
Chairs of minimum 12 mm diameter bars should be used.
Minimum diameter of dowel bars should be 12 mm
Lap slices should not be used for bar larger than 36 mm.
In steel reinforcement binding wire required is 8 kg per MT.
Lapping is not allowed for the bars having diameters more than 36 mm.
Minimum number of bars for a square column are 4 and for circular column are 6.
Longitudinal reinforcement should not be less than 0.8% and more than 6% of gross C/S.
Weight of rod per meter length = d2/162 where d is the diameter in mm
All reinforcement shall be free from mill scales, loose rust & coats of paints, oil or any other substances.
Main bars in the slabs shall not be less than 8 mm (HYSD) or 10 mm (Plain bars) and the distributors not less than 8 mm and not more than 1/8 of slab thickness.
In case of spacing of bars
Provide the diameter of the bar, if the diameter of the bar are equal.
Provide the diameter of the larger bar, if the diameter are unequal.
5mm more than the nominal maximum size of the coarse aggregate.
Provide the diameter of the bar, if the diameter of the bar are equal.
Provide the diameter of the larger bar, if the diameter are unequal.
5mm more than the nominal maximum size of the coarse aggregate.
Rolling margin in reinforcement steel | structural steel
What is rolling margin in reinforcement steel | structural steel
Rolling margin in reinforcement steel is percentage of deviation in Sectional weight of Reinforcement steel allowable as per IS codes.Reinforcement steel is extruded from a mould which is made for a particular size e.g 8mm Dia. When the mould is brand new,the sectional weight of 8mm steel extruded through mould would exact as per IS standard of lower than that.i.e. 0.395Kg per Metre or lessor. Mould gets little bigger after certain period of time or after certain quantity of production is taken from a particular mould. Now same 8mm dia bars extruded from the same mould will have more weight per Metre say 0.400Kg per Metre instead of 0.395 as per IS. That is more mass per Metre/Length is required for same length.This diviation in weight is defined in IS codes for different dia which is as under :-
Allowable deviation in rolling margin in reinforcement steel as per indian standards
8mm to 10mm +/- 7%
12mm to 16mm+/- 5%.
20mm and above+/-3%.
Rolling Margin is calculated as under :-
Total Weight of Bars (Dia wise ) / Total Running Metre of Bars = Actual Sectional Weight of bars.
Compare sectional weight with Standard IS Weight.
Measurement of bend dimension for stirrups and links
Description
Can any one derive the length calculation for the closed stirrups as per the IS-CODE 2502 1963
They have have given the closed stirrups (135degree hook) as
Total length = (2*(A+E))+(24*bardia )
explain how the 24 bardia is calculated
They have have given the closed stirrups (135degree hook) as
Total length = (2*(A+E))+(24*bardia )
explain how the 24 bardia is calculated
for 90 drgee bent (A+E-1/2R-d) cut length of bar . Generally the Deduction 90 degree bend only R calculated from IS 1786 cal9.4.1 Diameter of Mandrel
For stirrups having 3 90 drgee hook and 2 135 drgee hook No deduction for 135 drgee hook
For Example Beam size 200X600 cover to stirrups 25mm The size of stirrups required 150mmX550 Assume stirrups Dia 8mm Hook length 75mm
Cut length of stirrups
(150+550)X2-(10+8)X3+hook length (150)=1496mm
For Example Beam size 200X600 cover to stirrups 25mm The size of stirrups required 150mmX550 Assume stirrups Dia 8mm Hook length 75mm
Cut length of stirrups
(150+550)X2-(10+8)X3+hook length (150)=1496mm
Comments
Post a Comment