Handbook of Civil Engineering Calculations

(singke) #1

9012 - 59.16(4.73)


2
= 12,916 in
4
(53.76 dm
4
); Sb = 879 in
3
(14,406.8
cm^3 ); Sts = 570 in^3 (9342.3 cm^3 ).


  1. Transform the composite
    section, with cover plate
    included, to an equivalent
    homogeneous section of
    steel; compute the
    section moduli
    In accordance with AASHTO, the ef-
    fective flange width is 12(6.5) = 78 in
    (1981.2 mm). Using the method of an
    earlier calculation procedure, we see
    that when n = 30, y = 78/76.06 = 1.03
    in (26.162 mm); y'b = 19.42 + 1.03 =
    20.45 in (519.43 mm); y'ts = 17.92 -
    FIGURE 40. Transformed section. 1.03 = 16.89 in (429.006 mm); yt'c =
    16.89 + 6.50 = 23.39 in (594.106 mm);
    / = 12,802 + 9072 - 76.06(1.03)^2 =
    21,793 in^4 (90.709 dm^4 ); Sb = 1066 in^3
    (17,471.7 cm^3 ); Sts = 1,290 in^3
    (21,143.1 cm^3 ); Stc = 932 in^3 (15,275.5
    cm^3 ).
    When n = 10: y = 7.22 in (183.388 mm); yb = 26.64 in (676.66 mm); y'ts = 10.70 in
    (271.78 mm); J^ = 17.20 in (436.88 mm); I= 27,950 + 9191 - 109.86(7.22)^2 = 31,414 in^4
    (130.7545 dm^4 ); Sb=ll79 in^3 (19,320.3 cm^3 ); Sts = 2936 in^3 (48,121.0 cm^3 ); Stc = 1826
    in^3 (29,928. lcm^3 ).

  2. Transform the composite section, exclusive of the cover plate,
    to an equivalent homogeneous section of steel, and compute the
    values shown below
    Thus, when n = 30, yb = 23.78 in (604.012 mm); yt's = 12.06 in (306.324 mm); /= 14,549
    in^4 (60.557 dm^4 ); Sb = 612 in^3 (10,030.7 cm^3 ). When n = IQ 9 yb = 29.23 in (742.442 mm);
    y;s = 6.61 in (167.894 mm); /= 19,779 in


4
(82.326 dm
4
); Sb = 677 in
3
(11,096.0 cm
3
).


  1. Compute the dead load carried by the noncomposite member
    Thus,


Ib/lin ft N/m
Beam 150 2189.1
Cover plate 51 744.3
Slab: 0.54(6.75)(150) 547 7982.8
Haunch: 0.67(0.083)(150) 8 116.8
Diaphragms (approximate) 12 175.1
Shear connectors (approximate) 6 87.6
Total 774, say 780 11,383.2


  1. Compute the maximum dead-load moments
    Thus, M^L = (
    1
    / 8 )(0.250)(74.5)
    2
    (12) = 2080 in-kips (235.00 kN-m); M£L = (
    1
    A)
    (0.780)(74.5)
    2
    (12) = 6490 in-kips (733.24 kN-m).

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