* Minimum thickness of slab

Exterior Bay → t= L/24 = 15x12/24 = 7.5 in

Interior Bay → t= 1/28 = 8.5x12/28= 3.6 in

d=h-cover-0.5 (diam) = 7.5-0.75 -0.25 = 6.5"

Weight & slab = (7.5/12) x 150 pcf = 93.75 psf

Total Dead load = 93.75 + 70 163.75 psf

*

15

*

fo

8.5

8.5

Factored load: Wu= 1₁2DL + 1.6LL = 1.2 (163.75) + 1.6 (130) = 404.5 psf ~ 405 psf = 0.405 ksf

* Moment & Shear

SAP

using

IS/nYO

Determine the moment and shear using the ACI coefficient Method:

M=CM W₁12 and V = C, W₁1/2

Vmax at C= 1.15 (0.354)(11/2) = 2.24Kips

-ve Mmax at C = -(0.354)(11)²/10 = 4.3ft-kip

at B = +(0.354)(11)²/14 = 3.06ft-kip

+ve Mmax

●

CM.

CV.

A

1/24/14 %0 X₁

1.0

1.15 1.0

E

12

D

4

110

D

411

E

¼/16

CVE 313

Dr. Farid Abed/nL₂

Table 1: Dimensions and distributed loading for each group

Live

Load

psf

Group L₁

No.

(ft)

123456789

Bedroom

L3

L2

(ft) (ft)

5000

60

5000 60

14 17

13

13 15 17

15 15 16

14 15 16

14 14

4500

60

4000

60

16

4000 60

13 14 17

4500 60

15 16 17 4500 60

16 16 17 4000 60

15 15 16 4500 60

4555FHO65

fc

(ksi)

(psi)

3L2/4

7766ON6

Bathroom

120

120

120

120

130

130

130

130

110

Kitchen

Super

Imposed

Dead Load

psf

80

85

75

70

90

80

70

60

90

Living