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 19.22SP
Calculate the minimum main plate thickness for the joint of Problem 19.4 so that the plates, which are 7 in wide, will have a capacity equal to the shear capacity of the
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Design a single riveted lap joint to connect two mild steel plates, 14.5 mm thick, the joint being designed for maximum efficiency. The allowable stresses are σt = 55 N/mm2 , = 32.5 N/mm2 and σc = 63.5 N/mm2 . Also calculate the minimum force per pitch which will rupture the joint
Calculate the riveted joints, find its efficiency and draw a suitable sketch.
Thickness of the plates – t, mm is 12
Width of the plates – b, mm is 150
Load – P, kN is 20
Note. Choose a suitable material, determine its tearing stress. Take permissible shearing stress - teta=0.7sigmat ; Permissible crushing stress - sigmac=1.5sigmat ;
3. An M16 x 2 bolt with a slotted hex nut is used to clamp together two 25 mm steel plates. The joint
is assembled with a size 16 N washer underneath the bolt head. The bolt is a 5.8 class metric grade,
with a proof strength of 380 MPa. The joint is intended to be permanent. Determine a suitable length
for the bolt, rounded up to the nearest standard size in millimeters. Calculate the bolt stiffness, the
elongation of the bolt after the necessary preload is developed, and the bolt torque required to apply
the preload. Assume that coefficients of friction between the bolt, nut, and collar are all ~0.15. For
steel, use E = 207 GPa.
Standard threaded fastener lengths, dimensions of hexagonal nuts, and dimensions of metric plain
washers are provided in the textbook appendices.
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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- A 25 mm diameter steel bar is loaded in double shear until failure, the ultimate load is found to be 446 KN. If the allowable stress is to be based on a factor of safety 3, what must be the diameter of a pin designed for an allowable load of 26 KN in single shear be?arrow_forwardA wooden beam and a steel plate will be attached with three steel bolts. If the plate will be loaded with 120 kN, determine the necessary diameter for the bolts. Assume an ultimate shear stress corresponding to half the ultimate normal stress, for a tool steel, and a factor of safety of 3.25. Sound good material?arrow_forwardThis is a subject of Strength of MAterials, COmpute exactly without shortcut.Show FREE BODY DIAGRAM and COMPLETE SOLUTION thank youarrow_forward
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