Handbook of Civil Engineering Calculations

(singke) #1

  1. Determine whether the selected section is acceptable
    Compute the value OfKxL associated with a uniform-strength column, and compare this
    with the actual effective length. Thus, KxL = 1.73(16.2) = 28.0 > 27.6 ft (8.41 m). There-
    fore, the W8 x 40 is acceptable.


DESIGN OF AN AXIAL MEMBER


FOR FATIGUE


A web member in a welded truss will sustain precipitous fluctuations of stress caused by
moving loads. The structure will carry three load systems having the following character-
istics:


Force induced in member, kips (kN).
^ ^ ' No. of times
System Maximum compression Maximum tension applied

A 46(204.6) 18(80.1) 60,000
B 40(177.9) 9(40.0) 1,000,000
C 32(142.3) 8(35.6) 2,500,000


The effective length of the member is 11 ft (3.4 m). Design a double-angle member.

Calculation Procedure:


  1. Calculate for each system the design load, and Indicate the
    yield-point stress on which the allowable stress is based
    The design of members subjected to a repeated variation of stress is regulated by the
    AISC Specification, For each system, calculate the design load and indicate the yield-
    point stress on which the allowable stress is based. Where the allowable stress is less than
    that normally permitted, increase the design load proportionately to compensate for this
    reduction. Let + denote tension and - denote compression. Then


System Design load, kips (kN) Yield-point stress, kips/in^2 (MPa)

A -46-^2 / 3 (18) = -58(-257.9) 36(248.2)
B -40-^2 / 3 (9) = -^6(-204.6) 33(227.5)
C 1.5(-32 -^3 /4 x g) = -57 (-253.5) 33 (227.5)



  1. Select a member for system A and determine if it is adequate
    for system C
    From the AISC Manual, try two angles 4 x y/ 2 x % hi (101.6 x 88.90 x 9.53 mm), with
    long legs back to back; the capacity is 65 kips (289.1 kN). Then A = 5.34 in^2 (34.453
    cm
    2
    ); r = rx = 1.25 in (31.750 mm); KLIr = 11(12)71.25 = 105.6.
    From the Manual, for a yield-point stress of 33 kips/in
    2
    (227.5 MPa),/= 11.76
    kips/in^2 (81.085 MPa). Then the capacity P = 5.34(11.76) = 62.8 kips (279.3 kN) > 57
    kips (253.5 kN). This is acceptable. Therefore, use two angles 4 x 31/2 x % hi (101.6 x
    88.90 x 9.53 mm), long legs back to back.

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