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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Chapter 19, Problem 19.24SP
Rework Problem 19.23 changing the fasteners to six
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4. A solid shaft was made of alloy steel, it was welded to a flat plate by a
specific size of fillet weld. If a torque is applied to the shaft, calculate the size of
the weld joint.
Power
Torque
Yield stress Shear stress
Ultimate
stress (Mpa)
Compression
stress (Mpa)
Ductility (%)
(kw)
(Nm)
1000
(Mpa)
400
(Mpa)
160
280
30
Modulus of
Bending Stress Bending Moment
Factor of Safety
Elasticity
(GPa)
(Mpa)
220
280
120
4
Modulus of
Rigidity (pa)
70
Fracture stress
(Mpa)
250
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?
Q2/ Calculate the maximum axial stress for a 40-ft joint of 11.75-in, 60 lbm/ft, J-55 casing with short
threads coupling. Where the casing is subjected to 280000 lb, axial tension load in a portion of a
directional wellbore with dogleg severity of 2"/50 ft. In case of:
(1)Uniform contant between the casing and the borehole wall.
(2)Assuming contant between borehole wall and the casing only at the couplings,andF n+
(3)Calculate the joint setrength of the API round thread couplings.(oult = 75000 psi, wall thickness
0.489-in, Pc= 2660 psi. o yb-952000 lb, Fj-649 000 lbf).
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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- the shaft and Bols bolts. [Ans. 300 mm; 56 mm] 11. Design a bushed-pin type flexible coupling for connecting a motor shaft to a pump shaft for the following service conditions: Power to be transmitted = 40 kW; speed of the motor shaft= 1000 r.p.m. ; diameter of the motor shaft = 50 mm ; diameter of the pump shaft = 45 mm. The bearing pressure in the rubber bush and allowable stress in the pins are to be limited to 0.45 N/mm2 and 25 MPa respectively. [Ans. d, = 20 mm; n 6; d, = 40 mm; 1 152 mm] :1.1arrow_forward4. With tai yield strength of 393MP. JE the factor of safety is 4, calculate the allowable force E Strength of 210MPA. If the factor of safety is 4, calculate the key length. Note: fore circular shape welding A = 1.414 zhr J, = 2 nr F -045 930arrow_forwardDefine the term wide-flange sections?arrow_forward
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