Let us find out the cutting length of the haunch bar as shown below.
Given data:
Haunch = 300 × 300mm.
Wall thickness = 400mm.
Raft thickness = 400mm.
Clear cover = 50mm.
Haunch bar = 10mm @ 150 c/c.
Calculation:
The cutting length of the haunch bar
= [ (ab + bc +cd) - (2nos. × 45° bend)]
Where,
ab = cd =200mm. (given)
To find out the length of bc, first, we have to calculate the value of x.
x = [bk + ke.]
Let us draw a line at the clear cover, intersecting the haunch bar at s, as shown below.
From the above drawing,
ks = [wall thickness - (2 × cover)]
= [400mm. - (2 × 50mm.)]
= 300mm.
bk= ks = 300mm.
(as two angles are 45° )
Length of ke = [ (fg - y )+ ( raft thickness - cover)]
( from 2nd drawing.)
As haunch is 300 × 300mm., the length of side fg = 300mm.
Let us view the enlarged drawing so that we can calculate the value of y.
As the haunch bar is parallel to the haunch face, it makes a 45° angle at n & f.
mf = clear cover distance = 50mm.
As two angles are equal, side nm = side mf = 50mm.
By Pythagoras theorem,
nf2= (mf )2 +( nm )2
nf2= (50)2 +( 50 )2
nf = √5000
nf = 70.71mm.
y = [nf - cover]
= [70.71mm -50mm]
= 20.71mm.
As triangle nop is isosceles, side op = side no = y = 20.71mm.
Length of ke
= [ (fg - y )+ ( raft thickness - cover)]
= [ (300mm. - 20.71mm ) + ( 400mm. - 50mm.)]
= [279.29 + 350mm.]
= 629.29mm.
Length of x = [bk + ke.]
x = [300mm. + 629.29mm.]
= 929.29mm.
In triangle bec (2nd drawing), side be = side ec = x = 929.29mm.
By Pythagoras theorem,
bc2= (be)2 +(ec)2
bc2= (x)2 +( x)2
= (929.29)2 +( 929.29)2
bc = √ 1727159.81
= 1314.21mm.
The cutting length of the haunch bar
= [ (ab + bc +cd) - (2nos. × 45° bend)]
= [ (200mm. + 1314.21mm. + 200mm.) - (2nos. × 1 × 10mm.)]
= [ 1714.21mm. - 20mm.]
= 1694.21mm.
= 1.694m.
For BBS & cutting length of all types of structural members, click here.
Thank you for going through these calculation steps❤. Have a good day 😄.
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