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.
 

Rectangular plate with circular hole

 

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 × 106 mm4. 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.
 

Solution to Problem 335 | Flanged bolt couplings

Problem 335
The plate shown in Fig. P-335 is fastened to the fixed member by five 10-mm-diameter rivets. Compute the value of the loads P so that the average shearing stress in any rivet does not exceed 70 MPa. (Hint: Use the results of Prob. 332.)