Handbook of Civil Engineering Calculations

(singke) #1
From the AISC Manual for A440 steel:
Iff < 0.75 in (19.1 mm), J^ = 50 kips/in^2 (344.7
MPa).
If 0.75 < t < 1.5 in (38 mm), J^ = 46 kips/in^2
(317.1 MPa).
If 1.5 < t < 4 in (102 TXXOL) Jy = 42 kips/in
2
(289.5
MPa).


  1. Design the body of the member,
    using a trial thickness
    The Specification restricts the ratio w/t to a value
    of 8. Compute the capacity P of a % in (19.1-
    mm) eyebar of maximum width. Thus w = 8(%) =
    6 in (152 mm);/= 0.6(50) = 30 kips/in^2 (206.8
    MPa); P = 6(0.75)30 = 135 kips (600.5 kN). This
    is not acceptable because the desired capacity is
    175 kips (778.4 kN). Hence, the required thick-
    ness exceeds the trial value of^3 A in (19.1 mm).
    FIGURE 1. Eyebar hanger with * greater than % in (19.1 mm), the allowable
    stress at 1 is 0.60/;, or 0.60(46 kips/in^2 ) = 27.6
    kips/in^2 (190.3 MPa); say 27.5 kips/in^2 (189.6
    MPa) for design use. At 2 the allowable stress is
    0.45(46) = 20.7 kips/in^2 (142.7 MPa), say 20.5
    kips/in^2 (141.3 MPa) for design purposes.
    To determine the required area at 1, use the relation A 1 = PIf, where/= allowable
    stress as computed above. Thus, A 1 = 175/27.5 = 6.36 in
    2
    (4103 mm
    2
    ). Use a plate 6
    l
    /2 x 1
    in (165 x 25.4 mm) in which A 1 = 6.5 in^2 (4192 mm^2 ).

  2. Design the section through the pin hole
    The AISC Specification limits the pin diameter to a minimum value of 7w/8. Select a pin
    diameter of 6 in (152 mm). The bore will then be 6 !/32 in (153 mm) diameter. The net
    width required will be Pl(ft) = 175/[20.5(1.O)] = 8.54 in (217 mm); Dmin = 6.03 + 8.54 =
    14.57 in (370 mm). Set D = 143 A in (375 mm), A 2 = 1.0(14.75 - 6.03) = 8.72 in^2 (5626
    mm^2 ); A 2 IA 1 = 1.34. This result is satisfactory, because the ratio OfA 2 SA 1 must lie between
    1.33 and 1.50.

  3. Determine the transition radius r
    In accordance with the Specification, set r = D = 14% in (374.7 mm).


ANALYSIS OFA STEEL HANGER


A 12 x i/ 2 in (305 x 12.7 mm) steel plate is to support a tensile load applied 2.2 in (55.9
mm) from its center. Determine the ultimate load.

Calculation Procedure:


  1. Determine the distance x
    The plastic analysis of steel structures is developed in Sec. 1 of this handbook. Figure 2a
    is the load diagram, and Fig. 2b is the stress diagram at plastification. The latter may be
    replaced for convenience with the stress diagram in Fig. 2c, where T 1 = C; Pn = ultimate


Bore
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