Handbook of Civil Engineering Calculations

(singke) #1

  1. Design the weld required
    to develop the cover plate
    at each end
    Use fillet welds of E60 electrodes, placed
    along the sides but not along the end of the
    plate. The AISC Specification requires a
    minimum weld of^5 /ie in (7.9 mm) for a 1-in
    (25.4-mm) plate; the capacity of this weld is
    3000 Ib/lin in (525 N/mm). Then, length =
    97,400/3000 = 32.5 in (826 mm). However,
    the AISC requires that the plate be extended
    18 in (457 mm) beyond the theoretical cut-
    off point, thus providing 36 in (914 mm) of
    weld at each end. , %
    (b) Bending-moment diagram

  2. Design the intermittent weld
    The vertical shear at C is V 0 = R 0 - 5.3Ow = FIGURE 60. (a) Load diagram; (b) bend-
    64.45 - 5.30(3.303) = 46.94 kips (208.8 ing-moment diagram.
    kN); q = VQII = 46,940(0.33)712 = 1290
    Ib/lin in (225.9 N/mm). The AISC calls for a
    minimum weld length of I^1 A in (3.81 mm).
    Let s denote the center-to-center spacing. Then s = 2(1.5)3000/1290 = 7.0 in (177.8 mm).
    The AISC imposes an upper limit of 24 times the thickness of the thinner part joined, or


12 in (304.8 mm). Thus, (^5) max = 24(0.63) > 12 in (304.8 mm). Use a 7-in (177.8-mm) spac-
ing at the ends and increase the spacing as the shear diminishes.



  1. Design the shear connectors
    Use %-in (19.1-mm) studs, 3 in (76.2 mm) high. The design of the connectors is governed
    by the AISC Specification. The capacity of the stud = 11.5 kips (51.2 kN). From the AISC
    table, Vh = 453.4 kips (2016.7 kN). Total number of studs required - 2(453.4)/! 1.5 = 80.
    These are to be equally spaced.


DESIGN OF A CONCRETE JOIST


IN A RIBBED FLOOR


The concrete floor of a building will be constructed by using removable steel pans to form
a one-way ribbed slab. The loads are: live load, 80 lb/ft
2
(3.83 kPa); allowance for mov-
able partitions, 20 lb/ft
2
(0.96 kPa); plastered ceiling, 10 lb/ft
2
(0.48 kPa); wood floor
with sleepers in cinder-concrete fill, 15 lb/ft
2
(0.72 kPa). The joists will have a clear span
of 17 ft (5.2 m) and be continuous over several spans. Design the interior joist by the ulti-
mate-strength method, using// = 3000 lb/in
2
(20.7 MPa) mdfy - 40,000 lb/in
2
(275.8
MPa).


Calculation Procedure:


  1. Compute the ultimate load carried by the joist
    A one-way ribbed floor consists of a concrete slab supported by closely spaced members
    called ribs, or joists. The joists in turn are supported by steel or concrete girders that
    frame to columns. Manufacturers' engineering data present the dimensions of steel-pan
    forms that are available and the average weight of floor corresponding to each form.


(a) Load diagram
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