01-02ChapGere - 32 CHAPTER 1 Tension Compression and Shear...

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Problem 1.6-10 A flexible connection consisting of rubber pads (thickness t 5 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 g aver in the rubber if the force P 5 16 kN and the shear modulus for the rubber is G 5 1250 kPa. (b) Find the relative horizontal displacement d 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 5 9mm Length L 5 160mm Width b 5 80mm G 5 1250kPa P 5 16kN (a) S HEAR STRESS AND STRAIN IN THE RUBBER PADS (b) H ORIZONTAL DISPLACEMENT d 5 g aver t 5 (0.50)(9mm) 5 4.50mm g aver 5 t aver G 5 625 kPa 1250 kPa 5 0.50 t aver 5 P / 2 bL 5 8 kN (80 mm)(160 mm) 5 625 kPa P 2 P 2 P Thickness t Rubber pad Problem 1.6-11 Aspherical 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 t aver in the pin. (b) Determine the average bearing stress s 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 5 diameter of buoy 5 60in. T 5 tensile force in chain d p 5 diameter of pin 5 0.5in. t 5 thickness of shackle 5 0.25in. W 5 weight of buoy 5 1800lb g W 5 weight density of sea water 5 63.8lb/ft 3 F REE - BODY DIAGRAM OF BUOY T t d p T W F B F B 5 buoyant force of water pressure (equals the weight of the displaced sea water) V 5 volume of buoy F B 5g W V 5 4176lb 5 p d 3 6 5 65.45 ft 3 E QUILIBRIUM T 5 F B 2 W 5 2376lb (a) A VERAGE SHEAR STRESS IN PIN A p 5 area of pin (b) B EARING STRESS BETWEEN PIN AND SHACKLE A b 5 2 d p t 5 0.2500in. 2 s b 5 T A b 5 9500 psi t aver 5 T 2 A p 5 6050 psi A p 5 p 4 d p 2 5 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 5 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 5 250 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 5 100 mm. The force conditions between arm A and the lower flange are symmetrical with those given for arm B .
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This homework help was uploaded on 03/19/2008 for the course EM 319 taught by Professor Kennethm.liechti during the Spring '08 term at University of Texas.

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01-02ChapGere - 32 CHAPTER 1 Tension Compression and Shear...

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