Applied Statics and Strength of Materials (6th Edition)
6th Edition
ISBN: 9780133840544
Author: George F. Limbrunner, Craig D'Allaird, Leonard Spiegel
Publisher: PEARSON
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Textbook Question
Chapter 19, Problem 19.21SP
Rework Problem 19.20 changing the fasteners to
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A diagrammatic representation of a universal joint is shown, two yoke parts, the type being similar to Figs. 10.28 and 12.10, Text. The pin extensions have a diameter D = ¾ in.; a = 11/16 in., material of all parts is 4340, OQT Let N = 4 on ultimate stresses; n = 2400 rpm. Compute the safe torque for (a) shear of pins, (b) the pin extensions in bending, assuming that the load distribution is from zero at the outside pin ends to a maximum at the inside yoke surfaces, (c) an average bearing pressure on pins of 4 ksi. (d) What is the corresponding horsepower capacity?
OF 450 mm AND A WALL THICKNESS OF C6,25 mm IS FORMETD
A COMPRESSED AIR TANK HAVING AN INNER DIAMETER
BY WELDING TWO HEMISPHERES.
U. IF TIHE ALLOWABLE TENSILE STRESS IN THE STEEL IS 96 Mita,
WHAT IS THE MAXIMUM PERMISSIBLE AIR PRESSURE IH THE
TANK?
b. IF THE ALLOWABLE SHEAR STRESS IH THE STEEL IS 40 MPa,
WHAT IS THE MAXIMUM PERMISSIBLE PRESSURE IN THE TANK?
C. TEST ON THE WELDED SEAM SHOW THAT FAILURE OCCURS
WHEH THE TENSILE LOAD ON THE WELDS EXCEEDS
1.42 KN/mm OF WELD. IF THE REQUIRED FACTOR OF
SAFETY AGAINST FAILURE O THE WELD IS 2.5,
WHAT
Is THE MAXIMUM PERMISSIBLE PRESSURE IH THE TAHK?
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Chapter 19 Solutions
Applied Statics and Strength of Materials (6th Edition)
Ch. 19 - Prob. 19.1PCh. 19 - Rework Problem 19.1 assuming a bearing-type...Ch. 19 - Rework Problem 19.1 assuming a bearing-type...Ch. 19 - Compute the allowable tensile load for the...Ch. 19 - Rework Problem 19.4 assuming a bearing-type...Ch. 19 - Rework Problem 19.4 assuming that the bolts are 34...Ch. 19 - Select the number and arrangement of 34 in....Ch. 19 - Calculate the allowable tensile load for the...Ch. 19 - In the connection shown, 14 in. side and end...Ch. 19 - Design the fillet welds parallel to the applied...
Ch. 19 - A fillet weld between two steel plates...Ch. 19 - Design an end connection using longitudinal welds...Ch. 19 - Calculate the allowable tensile load for the butt...Ch. 19 - Calculate the allowable tensile load for the lap...Ch. 19 - Calculate the allowable tensile load for the butt...Ch. 19 - Rework Problem 19.10 assuming that both plates are...Ch. 19 - Rework Problem 19.12 assuming that the angle is an...Ch. 19 - Two ASTM A36 steel plates, each 12 in. by 12 in. ,...Ch. 19 - Rework Problem 19.20 changing the fasteners to 34...Ch. 19 - Calculate the minimum main plate thickness for the...Ch. 19 - A roof truss tension member is made up of 2L6412...Ch. 19 - Rework Problem 19.23 changing the fasteners to six...Ch. 19 - Determine the allowable tensile load that can be...Ch. 19 - The welded connection shown is subjected to an...Ch. 19 - In Problem 19.26, use a 38 in. fillet weld, change...
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- 7.22 Casing with an ID of 10.05 in. and a strength in tension of 450,000 lbf in the joints and 629,000 lbf in the pipe body is being suspended from the surface while being cemented. If the effective weight of the casing being sup- ported is 240,000 lbf and a safety factor of 1.4 is desired, how much surface pump pressure can be applied safely to displace the cement? The casing burst is rated at 3,130 psi (for zero axial stress).arrow_forwardFind the size of 14 bolts required for C.I.steam engine cylinder head. The diameter of the cylinder is 400 mm and the steam pressure is 0.12 N/mm.arrow_forwardSketch a V-grooved butt joint, and label all of the joint's dimensions.arrow_forward
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- (12) A turnbuckle connects two rods having %-16 UNC threads. The thread friction is 0.2. How much external torque is necessary to stress the rods 500lb if it is assumed that neither rod turns as the turnbuckle is tightened? The mean diameter of the threads is 0.335 in.arrow_forwardi need correct explanation my best wishes tonarrow_forwardQuo 2.11. Design the longitudinal and circumferential joint for a boiler whose diameteris 2.4 meter and is subjected to a pressure of 1 N/mm?. The longitudiral joint is a triple riveted butt joint with an efficiency of about 85% and the circumferentialjoint is a double riveted lap joint with an efficiency of nbout 70 %. The pitch in the outer rows of the rivets is to be double than in the inner rows and the width of the cover plate is unequal. The allowable stresses are: 0,=7 MPa, r = 56 MPa and o̟ = 120 MPa. Asgume that the resistance of rivets in double shear is 1.875 times that of single shear. Draw the complete joint.arrow_forward
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