text: # 1170365388 - 32 CHAPTER 1 Tension Compression and Shear...
  
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  Problem 1.6-10 A flexible connection consisting of rubber pads (thickness t 9 mm) bonded to steel plates is shown in the figure. The pads are 160 mm long and 80 mm wide. (a) Find the average shear strain aver in the rubber if the force P 16 kN and the shear modulus for the rubber is G 1250 kPa. (b) Find the relative horizontal displacement between the interior plate and the outer plates. 32 CHAPTER 1 Tension, Compression, and Shear P 2 P 2 P Rubber pad Rubber pad Section X-X X t = 9 mm X 80 mm t = 9 mm 160 mm Solution 1.6-10 Rubber pads bonded to steel plates Rubber pads: t 9 mm Length L 160 mm Width b 80 mm G 1250 kPa P 16 kN (a) S HEAR STRESS AND STRAIN IN THE RUBBER PADS (b) H ORIZONTAL DISPLACEMENT aver t (0.50)(9 mm) 4.50 mm g aver t aver G 625 kPa 1250 kPa 0.50 t aver P 2 bL 8 kN (80 mm)(160 mm) 625 kPa P 2 P 2 P Thickness t Rubber pad 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 posi- tioned 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. (a) (b) d Pin Shackle
  
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  Solution 1.6-11 Submerged buoy SECTION 1.6 Shear Stress and Strain 33 d diameter of buoy 60 in. T tensile force in chain d p diameter of pin 0.5 in. t thickness of shackle 0.25 in. W weight of buoy 1800 lb W weight density of sea water 63.8 lb/ft 3 F REE - BODY DIAGRAM OF BUOY T t d p T W F B F B buoyant force of water pressure (equals the weight of the displaced sea water) V volume of buoy F B W V 4176 lb d 3 6 65.45 ft 3 E QUILIBRIUM T F B W 2376 lb (a) A VERAGE SHEAR STRESS IN PIN A p area of pin (b) B EARING STRESS BETWEEN PIN AND SHACKLE A b 2 d p t 0.2500 in. 2 s b T A b 9500 psi t aver T 2 A p 6050 psi A p 4 d p 2 0.1963 in. 2 Problem 1.6-12 The clamp shown in the figure is used to support a load hanging from the lower flange of a steel beam. The clamp consists of two arms ( A and B ) joined by a pin at C . The pin has diameter d 12 mm. Because arm B straddles arm A , the pin is in double shear. Line 1 in the figure defines the line of action of the resultant horizontal force H acting between the lower flange of the beam and arm B . The vertical distance from this line to the pin is h 250 mm. Line 2 defines the line of action of the resultant vertical force V acting between the flange and arm B . The horizontal distance from this line to the centerline of the beam is c 100 mm. The force conditions between arm A and the lower flange are symmetrical with those given for arm B .
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