How to cut steel bars in construction projects.

1. Skills of steel bar blanking in actual construction: weld a bracket with thick steel pipe or channel steel thickened steel bar or angle steel, about 30 cm high, put it under the cutting machine, jam the caster of the cutting machine, raise the cutting machine, so that people will not lose their waist when working, so as to reduce fatigue strength, and make four or five iron benches, the volume is more than 60 cm square, the stool surface is made of iron plates, and the four legs are welded with steel bars.

Methods of steel bar blanking in actual construction;

Cutting length of beam and slab reinforcement = axial size of beam and slab-protective layer (generally 25)+ upper hook size 180 degree hook =6.25d 90 degree hook = 3.5d45 degree hook = = 4.9d.

Add salt again to remove the measurement difference: at 30 degrees, take 0.3d\ 45 degrees 0.5d\ 60 degrees 1d\ 90 degrees 2d\ 135 degrees 3d. If it is a general construction drawing, the distribution of reinforcement in the longitudinal stress of the slab/negative slab/beam can be calculated according to the above method. Frame reinforcement. If it is a plane construction drawing, please refer to 03g101-1b. The stirrup length is: outsourcing length+hook length -6d hook length 6 plus 100\8 plus 120+00 plus 1438.

Number of stirrups = beam member length -(25 protective layers) * 2/ stirrup spacing+1

(1) Calculation formula for blanking length of rectangular stirrups:

Stirrup blanking length = stirrup circumference+stirrup adjustment value (table 1), in which stirrup circumference =2 (outsourcing width+outsourcing length); Package width = b-2c+2d; Outsourcing length = h-2c+2d;

B×h= width× height of component section; C-protective layer thickness of longitudinal reinforcement; D—— stirrup diameter.

(2) Calculation example:

The section size of an anti-seismic frame beam is b×h=250mm×500mm, the reinforcement stirrup in the beam is φ6@ 150, and the thickness of the longitudinal reinforcement protective layer is c = 25 mm Find the blanking length of a stirrup. Solution: package width = b-2c+2d = 250-2× 25+2× 6 = 212 (mm) package length =h-2c+2d.

= 500-22× 25+2× 6 = 462mm

Stirrup blanking length = stirrup circumference+stirrup adjustment value =2 (outsourcing width+outsourcing length)+1 10 (adjustment value) = 2 (212+462)+10 =/kloc.