Eccentric Bolted Connections: Shear Only

Eccentric Bolted Connections: Shear Only. Ultimate Strength Analysis.

The major flaw in the analysis is the implied assumption that the fastener load–deformation relationship is linear and that the yield stress is not exceeded. Experimental evidence shows that this is not the case and that individual fasteners do not have a well-defined shear yield stress. The procedure to be described here determines the ultimate strength of the connection by using an experimentally determined nonlinear load–deformation relationship for the individual fasteners..

The ultimate strength of the connection is based on the following assumptions: l. At failure, the fastener group rotates about an instantaneous center (IC). 2. The deformation of each fastener is proportional to its distance from the IC and acts perpendicularly to the radius of rotation. 3. The capacity of the connection is reached when the ultimate strength of the fastener farthest from the IC is reached. (Figure 8.7 shows the bolt forces as resisting forces acting to oppose the applied load.) 4. The connected parts remain rigid. As a of the second assumption, the deformation of an individual fastener is r = —L-(0.34) where r = distance from the IC to the fastener max = distance to the farthest fmstener A = deformation of the farthest fastener at ultimate = 0.34 in. (determined experimentally) As with the elastic analysis, it is more convenient to work with rectangular com- ponents of forces, or Ry- R and Rx -ER where x and y are the horizontal and vertical distances from the instantaneous center to the fastener. At the instant of failure, equilibrium must be maintained, and the fol- lowing three equations of equilibrium will be applied to the fastener group (refer to Figure 8.7):.

The general procedure is to assume the location of the instantaneous center, then determine if the corresponding value of P satisfies the equilibrium equations. If so, this location is correct and P is the capacity of the connection. The specific procedure is as follows..

EXAMPLE 8.2. The bracket connection shown in Figure 8.8 must support an eccentric load consisting of 9 kips of dead load and 27 kips of live load. The connection was designed to have two vertical rows of four bolts, but one bolt was inadvertently omitted. If 7⁄8-inch-diameter Group A bearing-type bolts are used, is the connection adequate? Assume that the bolt threads are in the plane of shear. Use A36 steel for the bracket, A992 steel for the W6 × 25, and perform the ultimate strength analysis..

Solution.

Solution. For the holes nearest the edge, use 15/16 h 2 = 1.531 in. 2 To determine which component has the smaller bearing strength, compare the values of tFu (other variables are the same). For the plate, 5 (58) = 36.25 kips/in. 8 For the W6 x 25, — 29.58 kips/in. < 36.25 kips/ in. tFu = tfFu = 0.455(65) -.

Solution. The strength of the W 6 >< 25 will control. Rn = 1.2LctFu = = 54.34 kips 7 Upper Ilmlt — • • —2.4dtFu = 2.4 8 = 62.11 kips > 54.34 kips Use Rn For the other holes, use s = 3 in. Then, 15 s h in. 16 = 54.34 kips for this bolt. Rn = 1.2 LctFu = = 73.22 kips 2.4dtFu = 62.11 kips < 73.22 kips . Use Rn = 62.11 kips for these bolts..

Solution. Instead of incorporating a numerical value for Rult, we will use a unit strength and modify Equation 8.2 as.

Solution. TABLE 8.1 Origin at Bolt 1 Origin at IC Bolt 1 2 3 4 5 6 7 Sum x 0 3 0 3 o 3 O o 0 3 3 6 6 9 x 0.285 3.285 0.285 3.285 0.285 3.285 0.285 -3.857 -3.857 —0.857 —0.857 2.143 2.143 5.143 3.868 5.067 0.903 3.395 2.162 3.922 5.151 0.255 0.334 0.060 0.224 0.143 0.259 0.340 0.956 0.980 0.644 0.940 0.860 0.958 0.982 3.699 4.968 0.582 3.192 1.859 3.758 5.056 23.1120 0.071 0.636 0.203 0.910 0.113 0.802 0.054 2.78951.

Solution.

Solution. The nominal strength of one bolt, based on shear, is Rn = 28.86 kips. The nominal strength of the connection is.

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