and vertical components, respectively. Thus, Fx = MfS = 2500/146.3 = 17.09 kips
(76.0 kN); Fy = 40/15 = 2.67 kips (11.9 kN); F = (17.09^2 + 2.67^2 )^05 = 17.30 < 17.65.
Therefore, this is acceptable.
- Compute the stresses in the web plate at line 1
The plate is considered continuous; the rivet holes are assumed to be 1 in (25.4 mm) in di-
ameter for the reasons explained earlier.
The total depth of the plate is 51 in (1295.4 mm), the area and moment of inertia of the
net section are An = 0.416(51 -15x1)= 14.98 in^2 (96.6 cm^2 ) and/,, = (1/12)(0.416)(51)^3
- 1.0(0.416)(3510) = 3138 in^4 (130,603.6 cm^4 ).
Apply the general shear equation. Since the section is rectangular, the maximum
shearing stress is v = 1.5VIAn = 1.5(40)714.98 = 4.0 kips/in^2 (27.6 MPa). The AISC Speci-
fication gives an allowable stress of 14.5 kips/in^2 (99.9 MPa).
The maximum flexural stress is/= McIIn = 2500(25.5)13138 = 20.3 < 27 kips/in^2
(186.1 MPa). This is acceptable. The use of 15 rivets is therefore satisfactory.
- Compute the stresses in the rivets on line 2
The center of rotation of the angles cannot be readily located because it depends on the
amount of initial tension to which the rivets are subjected. For a conservative approxima-
tion, assume that the center of rotation of the angles coincides with the horizontal cen-
troidal axis of the rivet group. The forces are Fx = 2500/[2(146.3)] = 8.54 kips (37.9 kN);
Fy = 40/30 =1.33 kips (5.9 kN). The corresponding stresses in tension and shear are st =
FyIA = 8.54/0.6013 = 14.20 kips/in^2 (97.9 MPa); ss = Fy/A = 1.33/0.6013 = 2.21 kips/in^2
(15.2 MPa). The Specification gives ^allow = 28 - 1.6(2.21) > 20 kips/in^2 (137.9 kPa).
This is acceptable.
- Select the size of the connection angles
The angles are designed by assuming a uniform bending stress across a distance equal to
the spacing p of the rivets; the maximum stress is found by applying the tensile force on
the extreme rivet.
Try 4 x 4 x 3/4 in (102 x 102 x 19 mm) angles, with a standard gage of 21 X 2 in (63.5
mm) in the outstanding legs. Assuming the point of contraflexure to have the location
specified in the previous calculation procedure, we get c = 0.6(2.5 - 0.75) = 1.05 in (26.7
mm); M= 8.54(1.05) = 8.97 in-kips (1.0 kN-m);/= 8.97/[(^1 / 6 )(3)(0.75)^2 ] = 31.9 > 27
kips/in^2 (186.1 MPa). Use 5 x 5 x % in (127 x 127 x 22 mm) angles, with a 21 X 2 -Ui (63.5-
mm) gage in the outstanding legs.
- Determine the number of rivets required on line 3
The forces in the rivets above this line are shown in Fig 6a. The resultant forces are
FIGURE 6
Top of beam
