∑M

*

2,591"k

pc

= 53 - in3 (868.5X103 mm3)

⇒ Z req'd

≥

and;

∑M

*

49.3ksi

pb

Note: By inspection, shear does not govern.

(6) Second Floor Slab. Per Table 9.5(a) of ACI 318-95 the minimum thickness of a one way slab when

deflections are not computed is;

l/24

(one end continuous)

(governs)

l/28

(both ends continuous)

Therefore, for a 10-ft (3.05m) span; 10' 12"/1' /24 = 5" (127.0mm)

(

)

(7) Second Floor Longitudinal Beams. Use Fy = 36ksi (248.2MPa)

Strength requirements;

Note: Beam has continuous lateral support from attached roof deck.

wu = 1.2wDL+1.6wLL

where; wDL = (5/12)' 150pcf)10' 10psf(10' +(1psf+3psf+1psf)10'

(

+

)

Partition wt.

Suspended ceiling wt.

Beam Spacing

Miscellaneous wt.

Floor finish wt.

wDL = 775plf (11.30KN/m)

wLL = 40psf(10' = 400plf (5.83KN/m) (Per ANSI/ASCE 7-95 Table 4-1 Residential)

)

wu = 1.2(0.775klf)+1.6(0.400klf) = 1.57klf (22.90KN/m)

1.57klf (15' ) 2

= 44.2 ft - kips = 530in- kips (59.0KN/m)

Mu =

8

530in- kips (1000lb /k )

Mu

= 16.4 - in3 (268.7X103 mm3)

Z req 'd ≥

=

0.9(36,000psi)

w u L 1,570plf (15' )

= 12 k (53.4KN)

=

2

2

Deflection requirements;

5w L L4

5(0.40klf )(15' ) 4 (12"/1' )3

= 20.95 - in4 (8.72X106 mm4)

I req'd ≥

=

384E∆ allow

384(29,000ksi)0.75"

(8) Transverse Second Floor Beams. Use Fy = 36ksi (248.2MPa). Produce one design for the worst case

and use throughout. Worst case situation occurs at the interior bay.

Strength requirements;

Note: Beam has continuous lateral support from attached slab.

wu = 1.2wDL+1.6wLL

where; wDL = [(5/12)' 150pcf)+17plf(1/10' +12psf]15' 30plf

(

)

+

Self wt.

Tributary width

Concrete floor slab wt.

Partitions & Misc. wts.

Long beam wts.

H4-7

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