Let us now calculate the cutting length of the trapezium-shaped stirrup as shown below.
Given data:
The cutting length formula for trapezium stirrup
= [(a + 2b + c ) +(2nos × hook length) - {(4nos. × bend > 90°) + (1no.× bend < 90° )}]
I have given a trapezium stirrup drawing showing h, a, b, & c., where blue colored line is the centerline of the trapezium stirrup.
First, we will calculate the value of c,a, & b.
c = [column width -{( 2nos.×clear cover) - (2nos. ×1/2 × dia. of stirrup)}]
= [ 750mm - {( 2nos. × 40mm ) - ( 2nos. × 1/2 × 8mm )}]
= [750mm - { 80mm - 8mm}]
= [ 750mm - 88mm]
c = 662mm.
a = [x + {(2nos.×1/2 × dia.of main bar) + (2nos. ×1/2 ×dia. of stirrup.)}]
Here,
a = [x + {(2nos.×1/2 × dia.of main bar) + (2nos. ×1/2 ×dia. of stirrup.)}]
= [ 162mm +{( 2nos.× 1/2 × 20mm ) + ( 2nos. × 1/2 × 8mm )}]
= [ 162mm + { 20mm + 8mm }]
=[ 162mm + 28mm]
a = 190mm.
I have drawn the triangle part of the trapezium stirrup, for your clear understanding.
[ b is a hypotenuse of a triangle ]
Here,
y =[ x+ (1/2 × dia. of stirrup)]
= [162mm + (1/2 × 8mm.)]
y = 166mm.
h= [column depth - {(2nos ×clear cover) + (2nos.× 1/2 × dia. of stirrup)}]
=[450mm - {( 2nos.× 40mm ) + ( 2nos. ×1/2 × 8mm )}]
= [ 450mm - { 80mm + 8mm }]
= [ 450mm - 88mm ]
h= 362mm.
b =√( y )2 +( h )2
=√( 166 )2 +( 362 )2
= √27556 +131044
=√158,600
= 398.246mm.
The cutting length of trapezium stirrup
= [(a + 2b + c ) +(2nos × hook length) - {(4nos. × bend > 90°) + (1no.× bend < 90° )}]
= [ { 190mm + (2 × 398.246mm) + 662mm } + ( 2nos.× 10d ) - {(4nos. × 3d) + (1no.× 2d )}]
{Here, 10d is taken as hook length, 3d for the bend > 90°, 2d for bend < 90° & d is stirrup bar dia.}
= [ {190mm + 796.49mm + 662mm } + ( 2nos.× 10×8mm ) - {(4nos. × 3×8mm) + (1no.×2×8mm )}]
= [ 1648.49mm + 160mm - { 96mm + 16mm }]
= [ 1808.49mm - 112mm]
= 1696.49mm.
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