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01-02ChapGere.0002

# 01-02ChapGere.0002 - Line 1 in the figure defines the line...

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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.
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Unformatted text preview: 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 . Determine the average shear stress in the pin at C when the load P 5 18 kN. Arm A Arm B Line 2 Line 1 P C c Arm A h P A-PDF Split DEMO : Purchase from www.A-PDF.com to remove the watermark...
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