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Awesome engineering

الأربعاء، 13 أغسطس 2014

Design of slab forms



Design of slab forms can be summarized in the following design steps:
Step 1: Estimate design loads
Step 2: Determine sheathing thickness and and spacing of its supports (joist spacing)
Step 3: Determine joist size and spacing of supports (stringer spacing)
Step 4: Determine stringer size and span (shore spacing)
Step 5: Perform shore design to support stringers
Step 6: Check bearing stresses

Step 7: Design lateral bracing


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 مسألة في تصميم القوالب


طريقة و معادلات تصميم الشدة الخشبية

Design calculations:

1- design of decking


Weights:
O.W = 900 * 0.025 = 22.5 kg/m²
Weight of concrete = 25*25 = 625 kg/m²

Live loads ( ready mix with concrete pump) = 240 kg/m²


Total dead loads = 647.5 = 650 kg/m²
Live loads = 240 kg/m²
Total loads in sq. meter = 890 kg/m²


Moments = WS²/ 10
= 890 * .4 ²/10 = 14.24 kg.m

Max. shear force = .6 w s = .6*890*.4 = 213.6 kg

F1,2 max. stresses on outer fiber = M/I*Y = M/S where S = I/Y


,S =bh³/ (12*h/2) = bh²/ 6 = 100*2.5²/6 = 104 cm³

F max. = 14.24 * 10 ² /104 = 13.69kg/cm² O.K
Qmax. = 1.5 Q/A = 1.5*213.6 / (100*2.5) =1.28 kg/cm² O.K


2 – design of joists :


Weights:
O.W = 900 * 0.025 = 2.25 kg/m´
Weight of decking = 1.10*650*.40 = 286 kg/m´

Live loads = 240 kg/m²
( ready mix with concrete pump) = 1.10*240*.40 = 105.6 kg/m´

Total loads = 393 kg/m´

Mmax = Wl²/10
=393* .7²/10 = 19.3kg.m

Max. shear force = .6 w s = .6*393*.7 = 165kg

F1,2 max. stresses = M/I*Y = M/S where S = I/Y

,S =bh³/ (12*h/2) = bh²/ 6 = 100*2.5²/6 =41.6 cm³


F max. = 19.3 * 10 ² /41.6= 46.4 kg/cm² O.K


Qmax. = 1.5 Q/A = 1.5*165 / (12.5*2.5) =7.92kg/cm² taken 1"*5"


3- Design of wailing:


Weights:

O.W = 900 * 0.025 *.2 = 4.5 kg/m`

Loads = Rj = 1.10 *393 *.7 = 302 kg

Mmax = 45 kg.m

Max. shear force = 302 kg

F1,2 max. stresses = M/I*Y = M/S where S = I/Y

,S =bh³/ (12*h/2) = bh²/ 6 = 20²*2.5/6 = 166.6 cm³


F max. = 45 * 10 ² /166.66= 27 kg/cm² O.K


Qmax. = 1.5 Q/A = 1.5*302 / (25 *2.5) =7.25kg/cm² O.K



Design of prob:


load on prob :

Pp = 1.15 (2Rj + Ww*Lw)
= 1.15 (2*302 + 4.5*.7 ) = 699 kg


Fpb = 70 (1- Lb/80 bp) Lb = 185 cm

Fpb = 70 (1-185/(80*7.5)) = 48.4 kg/cm²

Area required = 699 /48.4 = 14.4 cm²


3"*3" prob is sufficient @ 700 mm in two direction O.k

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