Dear sir, How to calculate the average spacing of ties, if a column having length of 3 m. The spacing of ties in CR (i.e. 1 m) is 125 mm and in NR (i.e. 2 m) is 175 mm. Thanks
Divide the column length into three parts as l/3 or l/4 then the regions closer to joints are subjected to greater moment and torsion hence requires closer spaced stirrups to bind the longitudinal rebars together than at central portion if 2-legged ties is used then area of stirrups = 2*pie*8^/4 =100.53 .. considering the shear requirement from design as 350mm2/m the spacing required = (1000/350)*100.53 = 287.2 mm As per 26.5.3.2 clause of IS 456 the spacing of ties should be less than maximum of least column dimension 16 * dia of smallest longitudinal bar 300 mm
Dear sir,
How to calculate the average spacing of ties, if a column having length of 3 m. The spacing of ties in CR (i.e. 1 m) is 125 mm and in NR (i.e. 2 m) is 175 mm.
Thanks
Divide the column length into three parts as l/3 or l/4 then the regions closer to joints are subjected to greater moment and torsion hence requires closer spaced stirrups to bind the longitudinal rebars together than at central portion
if 2-legged ties is used then area of stirrups = 2*pie*8^/4 =100.53 .. considering the shear requirement from design as 350mm2/m the spacing required = (1000/350)*100.53 = 287.2 mm
As per 26.5.3.2 clause of IS 456
the spacing of ties should be less than maximum of
least column dimension
16 * dia of smallest longitudinal bar
300 mm
Why you didn't apply any wund load or seismic load?
lateral load not considered
HOLA NECESITO AYUDA EN ESTE PROGRAMA NO SE NADA MEIDEY
PLs stop music in background ,it distrcats focus
How to fine midline
select line element,right click,insert node,add midpoint
Thanks
Welcome
Thanks alot
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provide concrete 20 instead of 20000 kn/m2
Yes. I had entered *20 in kn/m2 it should have been 20.000 kn/m2 for M20 grade of concrete* as 1000 kn/m2 = 1 mpa
Por favor voy a llorar
auchilio