## Example

16 in. x 70 in. and two flange plates l1 in. x 20 in. for the girder cross-section.

Design bearing stiffeners for the plate girder of the preceding example for a factored end reaction of 260 kip.

Since the girder end is unframed, bearing stiffeners are required at the supports. The size of the stiffeners must be selected to ensure that the limit states of local buckling, compression, and bearing are not violated.

Limit state of local buckling: Refer to Figure 4.13, try bst = 8 in. To avoid problems with local buckling bst/2tst must not exceed 0.56^/(E/Fy) = 15.8. Therefore, try tst = 2 in. So, bst/2tst = 8, which is less than 15.8.

Plate girder flange

Plate girder flange

Limit state of compression:

Aeff = 2(bsttst) + 12tW = 2(8)(0.5) + 12(5/16)2 = 9.17 in.2 Ist = tst(2bst + i„)3/12 = 0.5[2(8) + 5/16]3/12 = 181 in.4 rst = V(IjAs) = P(181/9.17)= 4.44 in. Kh/rst = 0.75(70)/4.44 = 11.8

1c = (Kh/prst)P(Fy/E) = (11.8/3.142)P(36/29,000) = 0.132

and from Equation 4.17

fcPn = 0 .85(0 . 6581)FyAst = 0 . 85(0. 658)°' 1322 (36)(9.17) = 279 kip

Since f cPn > 260 kip, the design is satisfactory for compression.

Limit state of bearing: Assuming there is a 4-in. weld cutout at the corners of the bearing stiffeners at the junction of the stiffeners and the girder flanges, the bearing area for the stiffener pairs is: Apb = (8 - 0.25)(0.5)(2) = 7.75 in.2. Substituting this into Equation 4.98, we have fRn = 0.75(1.8)(36)(7.75) = 377 kip, which exceeds the factored reaction of 260 kip. So bearing is not a problem.

Use two 1 in. x 8 in. plates for bearing stiffeners.

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