# 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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The text passage contains two distinct problems, each with a question and a solution. The first problem deals with the shear strain and displacement in rubber pads, while the second problem involves the shear stress and bearing stress in a pin connecting a submerged buoy to a chain.
multiple_questions_with_answers
Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
- NO_QUESTION
- SINGLE_QUESTION
- SINGLE_QUESTION_WITH_ANSWER
- MULTIPLE_QUESTIONS
- MULTIPLE_QUESTIONS_WITH_ANSWERS
system
# 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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### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern
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The text passage contains two distinct problems, each with a question and a solution. The first problem deals with the shear strain and displacement in rubber pads, while the second problem involves the shear stress and bearing stress in a pin connecting a submerged buoy to a chain.

TYPE: MULTIPLE_QUESTIONS_WITH_ANSWERS