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TABLE 22.1 Allowable Stresses
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TABLE 22.1 Allowable Stresses (Continued)
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AS =
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0.9743 d 4 n
(22.4) Metric: AS = (d 0.9382P)2 4
Here is an example based on Fig. 22.2. The bolts are ASTM A325 steel, m = 5 bolts, F = 38 250 lb (170.1 kN), d = 3 4 in (19.1 mm), b = 2 (one through the body of each bolt, one through the threads), and n = 12 threads per inch (2.12 mm per thread). The total cross-sectional area through the bodies of all five bolts and then through the threads is 5AB = 5AS = 5 (0.75)2 = 2.209 in2 (1425 mm2) 4 5 0.9743 0.75 4 12
= 1.757 in2 (1133 mm2)
The shear stress in each bolt will be = F 38 250 = = 9646 psi (66.5 MPa) AT 2.209 + 1.757
which is well within the shear stress allowed for A325 steel bolts (Table 22.1). Tensile Stress in the Plate. To compute the tensile stress in the plates (we will assume that these are made of A36 steel), we first compute the cross-sectional area of a row containing the most bolts. With reference to Figs. 22.2 and 22.3, that area will be
FIGURE 22.2 Shear joint example. The joint and splice plates here are each 3 4 in (19.1 mm) thick. Dimensions given are in inches. To convert to millimeters, multiply by 25.4.
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BOLTED AND RIVETED JOINTS 22.11
BOLTED AND RIVETED JOINTS
FIGURE 22.3 Tensile failure of the splice plates. Tensile failure in the plates occurs in the cross section intersecting the most bolt holes.
A = 0.75(1.5) + 0.75(3) + 0.75(1.5) = 4.5 in2 (2903 mm2) The stress in two such cross sections (there are two splice plates) will be = F 38 250 = = 4250 psi (29.3 MPa) A (4.5)2
These plates will not fail; the stress level in them is well within the allowable tensile-stress value of 21.6 kpsi for A36 steel. In some joints we would want to check other sections as well, perhaps a section in the splice plate. Bearing Stresses on the Plates. If the fasteners exert too great a load on the plates, the latter can be deformed; bolt holes will elongate, for example. To check this possibility, the designer computes the following (see Fig. 22.4): B = F mdlG
For our example, lG = 2.25 in (57.2 mm), m = 5, and d = 0.75 in (19.1 mm). Then B = 38 250 = 4533 psi (31.3 MPa) 5(0.75)(2.25)
FIGURE 22.4 The bearing area of a bolt. The dimensions given are those used in the example in the text for the joint shown in Fig. 22.2. Dimensions are in inches. Multiply by 25.4 to convert to millimeters.
Note that the allowable bearing stresses listed in Table 22.1 are greater than the allowable shear stresses for the same plate material. Tearout Stress. Finally, the designer should determine whether or not the fasteners will tear out of a joint plate, as in the lap joint shown in Fig. 22.5. In the example shown there are six shear areas. The shear stress in the tearout sections will be = 100 000 = 11 111 psi (76.6 MPa) 6(0.75)(2)
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