Handbook of Civil Engineering Calculations

Calculation Procedure: Compute the required area of the base plate; establish the plate dimensions Refer to the base-plate diag ...

Recording the allowable stresses and modular ra- tio by using the ACI Code, we get p = 750 lb/in 2 (5170 kPa) and n = 9. From th ...

(b) Forces and moment (d) Force on weld FIGURE 13. (a) Stresses; (b) forces and moment; (c) stresses on projection; (d) force on ...

FIGURE 14. Grillage under column. mum bending moment occurs at the center of the span; its value is M = P(A - a)/8 =fS; therefor ...

The pressure under the plate - 2790/[34(3I)] = 2.65 kips/in^2 (18.3 MPa). For a 1-in (25.4- mm) width of plate, M= V2(2.65)/(l . ...

Note Data recorded in following order: shear, end moments, axial force. FIGURE 15. Wind-stress analysis by portal method. ...

The shear factor equals the ratio of the average width of the adjacent aisles to the total width. Or, line A 9 15/75 = 0.20; lin ...

Note Data recorded in following order: shear, end moments, axial force. FIGURE 17. Wind-stress analysis by cantilever method. ...

considers that the bent behaves as a vertical cantilever. Consequently, the direct stress in a column is directly proportional t ...

WIND-STRESS ANALYSIS BY SLOPE-DEFLECTION METHOD Analyze the bent in Fig. 180 by the slope-deflection method. The moment of inert ...

applied solely at its ends. The sign convention is as follows: an end moment is positive if it is clockwise; an angular displace ...

Compute the shear in each member by analyzing the member as a free body The shear is positive if the transverse forces exert a ...

TABLE 1 Calculation of Wind Drift Member 7, in^4 (cm^4 ) L, ft (in) M 69 ft-kips (kN-m) me, ft-kips (kN-m) MjTi 6 LII A-U-U 1,50 ...

FIGURE 21. Bending-moment diagrams. A = 19.25(12)^2 /[3(29)(10)^3 ] = 0.382 in (9.7 mm). For dimensional homogeneity, the left s ...

sion to 300, and another section provides an allowable stress of 22 kips/in^2 (151.7 MPa). Thus, L^2 = 142 + 122 = 340 ft^2 (31. ...

Manual, the allowable stress corresponding to this ratio is fc = \.661fb - 8640 - \(fb - 12,950)g/f]/15 = 1.667(20,000) - 8640 - ...

(3.96 m). If the yield-point stress is 33,000 lb/in 2 (227.5 MPa), compute the allowable unit load for this member and the corre ...

3.13(3.40 - 2.77) 2 0.581(3.40 - 0.053) 2 21.53 in 4 (896.15 cm 4 ). Then Sx = 21.53/4.60 = 4.68 in 3 (76.69 cm 3 ). This value ...

SECTION 2 REINFORCED AND PRESTRESSED CONCRETE ENGINEERING AND DESIGN MAX KURTZ, P.E. Consulting Engineer TYLER G. HICKS, P.E. In ...

DESIGN OF COMPRESSION MEMBERS BY ULTIMATE- STRENGTH METHOD Analysis of a Rectangular Member by Interaction Diagram Axial-Load Ca ...