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Mechanics of Materials – MAE 243 (Section 002) Spring 2008 Dr. Konstantinos A. Sierros Problem 1.6-5 The connection shown in the figure consists of five steel plates, each 3⁄16 in thick, joined by a single 1⁄4-in. diameter bolt. The total load transferred between the plates is 1200 lb, distributed among the plates as shown. (a) Calculate the largest shear stress in the bolt, disregarding friction between the plates. (b) Calculate the largest bearing stress acting against the bolt. Problem 1.6-11 A spherical fiberglass buoy used in an underwater experiment is anchored in shallow water by a chain [see part (a) of the figure]. Because the buoy is positioned just below the surface of the water, it is not expected to collapse from the water pressure. The chain is attached to the buoy by a shackle and pin [see part (b) of the figure]. The diameter of the pin is 0.5 in. and the thickness of the shackle is 0.25 in. The buoy has a diameter of 60 in. and weighs 1800 lb on land (not including the weight of the chain). (a) Determine the average shear stress aver in the pin. (b) Determine the average bearing stress b between the pin and the shackle. 1.7 Allowable stresses and allowable loads • The principal design interest is strength • Strength is the capacity of the object to support or transmit loads • The actual strength of a structure must exceed the required strength • Factor of safety must be greater than 1 if failure is to be avoided • Factors of safety from slightly above 1 to as much as 10 are used Factor of safety (n) = Actual strength / Required strength 1.7 Allowable stresses and allowable loads In aircraft design we use the term margin of safety rather than factor of safety Margin of safety = Factor of safety - 1 When the margin of safety is reduced to zero or less, the structure (presumably) will fail Allowable stresses Allowable stress must not be exceeded anywhere in the structure Allowable stress = Yield strength / Factor of safety σallow = σY / n1 tension τallow = τY / n2 shear OR… Allowable stress = Ultimate strength / Factor of safety tension σallow = σU / n3 shear τallow = τU / n4 Allowable loads • Allowable load is also called permissible load or the safe load Allowable load = (Allowable stress) (Area) • For bars in tension or compression: Pallow = σallow A • For pins in direct shear: Pallow = τallow A • Permissible load based upon bearing: Pallow = σb A 1.8 Design for axial loads and direct shear • When designing a structure, we must determine the properties of the structure so that the structure can support the loads and perform its functions FIG. 1-25 Bolted connection in which the bolt is loaded in single shear Steel eyebar subjected to tensile loads P FIG. 1-3 Bolted connection in which the bolt is loaded in double shear FIG. 1-24 Copyright 2005 by Nelson, a division of Thomson Canada Limited Copyright 2005 by Nelson, a division of Thomson Canada Limited Copyright 2005 by Nelson, a division of Thomson Canada Limited • Knowing the loads to be transmitted in the structure and the allowable stresses we can calculate the required area of members Required area = (Load to be transmitted) / (Allowable stress) Please study example 1.8 (pages 46-48) • Next time we will discuss about it. Therefore, you need to make sure that you have comments/questions ready for discussion Example 1-8. Two-bar truss ABC supporting a sign of weight W FIG. 1-33 Copyright 2005 by Nelson, a division of Thomson Canada Limited Monday 4 February 2008 during class… Quiz covering Chapter 1 Duration: 20 mins Solve 1 out of 2 Questions