Handbook of Civil Engineering Calculations

2. Determine the modified yield stress and modulus of elasticity
   Determine Fmy and Em:
   Ar Ac
   Fmy = Fy + clFyr — +c 2 f^—-
   AS As
   where Ar = the cross-sectional area of four No. 7 longitudinal bars = 4 x 0.6 in^2 = 2 4 in^2
   (15.5 cm^2 )
   A 5 = cross-sectional area of W8 x 40 = 11.7 in^2 (75.5 cm^2 )
   A 0 = 16 in x 16 in - (11.7 in^2 + 2.4 in^2 ) = 242 in^2 (1561 cm^2 )

For concrete-encased shapes, C 1 = 0.7 and C 2 = 0.6.

Fm, = 36 ksi + 0.7 x 55 ksi x-j^5_^2 4 in^2                                 242 + 0.6 x 3.5 ksi x ^B_ in^2

= 87.3 ksi (601.5 MPa)

Em = E + c£e^-

AS

where C 3 = 0.2 for concrete-encased shapes

EC = w^1 -^5 V^ = 1451 -^5 V3^5 = 3267 ksi (24,577 MPa) for 3.5-ksi normal-weight

(145 lb/ft^3 ) (2320 kg/cu m) concrete

Em = 29,000 ksi + 0.2 x 3267 ksi x 242 in^2 /! 1.7 in^2 = 42,513 ksi (292,915 MPa)

The modified radius of gyration

rm = >>(W8 x^4 °) ^ 0.3 x 16 in (overall dimension)

= 2.04 in > 4.80 in (12.2 cm)

= 4.80 in (12.2 cm)

The slenderness parameter

H JE^

c R

n* V Em

= 15.0 ft x 12 in/ft / 87.3 ksi

4.8OmX 17 V^42 >513ksi

The critical stress

Fcr = (0.658*c^2 )F^

= 0.658^^0 -^54 )^2 x 87.3 ksi = 77.2 ksi (531.9 MPa)

3. Compute the design compressive strength
   The design compressive strength

JJPn = Wf 0 ,

= 0.85 x 11.7 in^2 x 77.2 kips/in^2 (531.9 MPa)

= 768 kips (5292 MPa)