خطوات تصميم الدرج--

من طرف qaisalkurdy السبت أغسطس 25, 2007 5:04 am

خطوات تصميم الدرج stair design

stair design

length of horizontal projection=2.4m

height of rise=15cm

number of rise=H/15

number of rise=300/15=20

use two flight:

each flight contain 10 Rise

number of Go=number of rise-1

#G=#R-1

#G=10-1=9

space between flight=20cm

length of step=(2.6-.2)/2=1.2m

length of flight in horizontal projection

length=2.7+.5*1.2+.5*1.3

length=3.95m

hmin=Ln/33

hmin=3950/33=119.7mm

use h=150mm

θ=tan-1(R/G)

θ=tan-1(15/30)

θ=26.565

cosθ=.8944

Load calculation:

Dead load from:

1-slab =.15*1*1*24=3.6KN/m2

2-step=3.33*(0.5*0.15*.3)*24=1.80KN\m2

3-plaster=.02*1*1*21=.03*21=.42KN\m2

4-tile=1.45*1*0.03*22=.957

5-sand=.03*1*1*18=0.56KN\m2

∑DL=6.36KN\m2

dead load in horizontal projection ∑DL=6.36/cos8944

∑DL=11.37KN\m2

LL=3KN\m2

Wu=1.4DL+1.7LL

Wu=1.4*11.37+1.7*3

Wu=20.87KN\m2

Wu=20.87*1=20.87KN/m

Mu=WuL2/8

Mu=20.87*3.952/8

Mu=40.71KN.m

Mu=Φbd2f’cω(1-0.59ω)

ω(1-0.59ω)=Mu/(Φbd2f’c)

ω(1-0.59ω)=40.71/(0.9*1000*1202*25)

ω(ω1-0.59ω)=.1287

ω=0.1403

ρ=ωf’c/fy

ρ=0.1403*25/420

ρ=8.35*10-3

ρmin<ρ<ρmax

1.4/fy<ρ<.75*.85β1*(f’c/fy)*(600/(600+fy)

1.4/420<ρ<.75*.85*.85(25/420)*(600/(600+420))

.0033<.00835<.0189...ok

As=ρbd

As=.00835*1000*120

As=1002mm2/m

use 7Φ14/m

in other direction use minimum reinforcement

As=0.0018bh

As=.0018*1000*150

As=270mm2/m

use 3Φ12/m

landing design

L2=2.6m

L1=1.3

L2/L1=2.6/1.3≥2 one way solid slab

hmin=Ln/33

hmin=2600/33

hmin=79mm

use h=15cm

load calculation

1-slab =.15*1*1*24=3.6KN/m2

2-plaster=.02*1*1*21=.03*21=.42KN/m2

3-tile=1.45*1*0.03*22=.957

4-sand=.03*1*1*18=0.56KN/m2

5-reaction from flight=wL/2=6.36*3.95/2=12.56KN/m2

∑DL=18.1KN/m2

reaction from flight=wL/2=3*3.95/2=5.93KN/m

LL=3KN/m2

∑LL=8.93KN/m

Wu=1.4DL+1.7LL

Wu=1.4*18.1+1.7*8.93

Wu=40.52KN/m2

Mu=WuL2/12

Mu=38*1.32/12

Mu=5.71KN.m

Mu=Φbd2f’cω(1-0.59ω)

ω(1-0.59ω)=Mu/(Φbd2f’c)

ω(1-0.59ω)=5.71*106/(.9*1000*1202*25)

ω(1-0.59ω)=0.0176

ω=0.018

ρ=ωf’c/fy

ρ=0.017*25/420

ρ=1.07*10-3

ρmin<ρ<ρmax

1.4/fy<ρ<.75*.85β1*(f’c/fy)*(600/(600+fy)

1.4/420<ρ<.75*.85*.85(25/420)*(600/(600+420))

but

ρ<ρmin

use ρ=ρmin=.0033

As=ρbd

As=.0033*1000*120=396mm2/m

use 4Φ12

use minimum reinforcement in other direction

As=0.0018*b*h

As=0.0018*1000*150

As=270mm2/m

use 4Φ10/m