# plate

## 240 How to locate the centroid of metal plate with circular hole

**Problem 240**

The shaded area in Fig P-240 represents a steel plate of uniform thickness. A hole of 4-in. diameter has been cut in the plate. Locate the center of gravity the plate. *Hint:* The weight of the plate is equivalent to the weight of the original plate minus the weight of material cut away. Represent the original plate weight of plate by a downward force acting at the center of the 10 × 14 in. rectangle. Represent the weight of the material cut away by an upward force acting at the center of the circle. Locate the position of the resultant of these two forces with respect to the left edge and bottom of the plate.

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## Solution to Problem 597 | Spacing of Rivets or Bolts in Built-Up Beams

**Problem 597**

A plate and angle girder similar to that shown in Fig. 5-32 is fabricated by riveting the short legs of four 125 × 75 × 13 mm angles to a web plate 1000 mm by 10 mm to form a section 1020 mm deep. Cover plates, each 300 mm × 10 mm, are then riveted to the flange angles making the overall height 1040 mm. The moment of inertia of the entire section about the NA is *I* = 4770 × 10^{6} mm^{4}. Using the allowable stresses specified in Illustrative Problem 591, determine the rivet pitch for 22-mm rivets, attaching the angles to the web plate at a section where *V* = 450 kN.

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## Solution to Problem 335 | Flanged bolt couplings

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## Solution to Problem 334 | Flanged bolt couplings

**Problem 334**

Six 7/8-in-diameter rivets fasten the plate in Fig. P-334 to the fixed member. Using the results of Prob. 332, determine the average shearing stress caused in each rivet by the 14 kip loads. What additional loads P can be applied before the shearing stress in any rivet exceeds 8000 psi?

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## Solution to Problem 333 | Flanged bolt couplings

**Problem 333**

A plate is fastened to a fixed member by four 20-mm-diameter rivets arranged as shown in Fig. P-333. Compute the maximum and minimum shearing stress developed.

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