Handbook of Civil Engineering Calculations

DESIGN OF REINFORCEMENT IN A RECTANGULAR BEAM OF GIVEN SIZE A rectangular beam of 4000-lb/in 2 (27,580-kPa) concrete has a width ...

DESIGN OFA RECTANGULAR BEAM A beam on a simple span of 13 ft (3.9 m) is to carry a uniformly distributed load, exclu- sive of it ...

m). Using^' = 3000 lb/in^2 (20,685 kPa) and an allowable stress^, in the stirrups of 20,000 lb/in^2 (137,900 kPa), design web re ...

Select the stirrup size Use the method given earlier in the ultimate-strength calculation procedure to select the stirrup size, ...

Alternatively, calculate the allowable bending moment by assuming that the flange extends to the neutral axis Then apply the ne ...

The beam size is slightly deficient with re- spect to balanced design, and the concrete will therefore be stressed to capacity u ...

Calculation Procedure: Identify the controlling stress Thus, M= (^1 / 8 )(4100)(22)^2 (12) = 2,977,000 in-lb (336,341.5 N-m). F ...

steel, the ACI Code provides that "in doubly reinforced beams and slabs, an effective modular ratio of 2n shall be used to trans ...

FIGURE 19 Construct the banding-moment diagram Apply the ACI equation for maximum midspan moment. Refer to Fig. 19c: M 1 = (Ys) ...

Calculate the moment of inertia of the transformed cracked section at the center Referring to Fig. I9e and assuming tentatively ...

Calculate the moment of inertia of the transformed cracked section at the center Referring to Fig. I9e and assuming tentatively ...

(a) Section (b) Strains (c) Stresses (d) Resultant forces FIGURE 20 Calculation Procedure: Compute the value of c associated wi ...

c = 18/0.85 = 21.8 in (537.972 mm);/* = €<£& - d)lc = 87,000(21.18 - 15.5)721.18 = 23,300 lb/in^2 (160,653.5 kPa); F 0 = ...

FIGURE 21. Interaction diagram. 1,596,000/193,400 = 8.25 in (209.55 mm). This result discloses that an eccentricity of 9.2 in (2 ...

ALLOWABLE ECCENTRICITY OF A MEMBER The member analyzed in the previous two calculation procedures is to carry an ultimate longit ...

DESIGN OFA SPIRALLY REINFORCED COLUMN A short circular column, spirally reinforced, is to support a concentric load of 420 kips ...

ANALYSIS OFA RECTANGULAR MEMBER BY INTERACTION DIAGRAM A short tied member having the cross section shown in Fig. 22 is to resis ...

FIGURE 23. Interaction diagram. and the latter on an uncracked section. The subscript b as used by the ACI Code in the present i ...

Mb — = QA3pgmDs + O. I4t (47 a) Pb For symmetric tied columns: Mb — = d(Q.61Pgm + 0.17) (4Ib) Pb For unsymmetric tied columns: M ...

Calculation Procedure: Evaluate P when e = IO in (254 mm) As the preceding calculations show, the eccentricity corresponding to ...