Re analysis the designed interior panel by using ACI Coefficient Method for:

1- calculate allowable live load.

2- sketch the designed reinforcement.

3- check the shear by ACI method.

An edge slab (4.8 m * 6.0 m) has thickness (135 mm) , at short span the reinforcement is Ø 10 at 175 mm at the positive region and 2 Ø 10 mm at 350 mm (bent) + Ø 10 at 350 mm (added) at negative regions. at long span the reinforcement is Ø 10 at 200 mm at positive region and 2 Ø 10 at 400 mm + Ø 10 at 450 mm over the continuous edge.

Take the following :

- cover concrete 20 mm

- cylindrical stress  0.0225 KN/mm²

- yield strength for the second material is 0.375 KN/mm²

- Ƴc = 2.5 KN/m³

- added dead load = 0.85 KN/m³

Transcribed Image Text:

Direct design method (D.D.M) * check the restrictions:- 1- No. of span 23 (for each direction) long span 2- short span = s2 3. L-S s 4- S2 D4 5- Shift of column s 10% span 6- gravity load is applied only. 7- successive span in each not differ by 8- stiffness for tow beam in two perpendicular direction al• (L2) 0.2 < < 5 a2 • (L1) * In 2 0.65 L1 Mo = for each directiori -Mu 0.65 M.. +Mu 0.35 M. bh I beam am I slab = + Ad? Tslab 12: I beama 12 *% of the interior negative moment resisted by column strip :- 12 0.5 %75 %75 %90 S49 %75 *% of the exterior positive moment resisted by column strip :- 12 05 60 660 al 75 a 21 に 云 Direct design method (D.D.M) * check the restrictions:- 1- No. of span 23 (for each direction) long span 2- short span = s2 3. L-S s 4- S2 D4 5- Shift of column s 10% span 6- gravity load is applied only. 7- successive span in each not differ by 8- stiffness for tow beam in two perpendicular direction al• (L2) 0.2 < < 5 a2 • (L1) * In 2 0.65 L1 Mo = for each directiori -Mu 0.65 M.. +Mu 0.35 M. bh I beam am I slab = + Ad? Tslab 12: I beama 12 *% of the interior negative moment resisted by column strip :- 12 0.5 %75 %75 %90 S49 %75 *% of the exterior positive moment resisted by column strip :- 12 05 60 660 al 75 a 21 に 云

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